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ICSE Class X Board Exam 2024 : History and Civics

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Ayush Saha
St. Lawrence High School, Kolkata
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Lakshya JEE AIR (2025) 1 ELECTROSTATICS ELECTROSTATICS ELECTRIC CHARGE Charge is the property associated with matter due to which it produces and experiences electrical and magnetic effects. The excess or deficiency of electrons in a body gives the concept of charge. Types of charge : (i) Positive charge : It is the deficiency of electrons as compared to proton. (ii) Negative charge : It is the excess of electrons as compared to proton. SI unit of charge : ampere second i.e. Coulomb, Dimension : [A T] Practical units of charge are ampere hour (=3600 C) and faraday (= 96500 C) SPECIFIC PROPERTIES OF CHARGE Charge is a scalar quantity : It represents excess or deficiency of electrons. Charge is transferable : If a charged body is put in contact with an another body, then charge can be transferred to another body. Charge is always associated with mass Charge cannot exist without mass though mass can exist without charge. So the presence of charge itself is a convincing proof of existence of mass. In charging, the mass of a body changes. When body is given positive charge, its mass decreases. When body is given negative charge, its mass increases. Charge is quantised The quantization of electric charge is the property by virtue of which all free charges are integral multiple of a basic unit of charge represented by e. Thus charge q of a body is always given by q = ne n = positive integer or negative integer The quantum of charge is the charge that an electron or proton carries. Note : Charge on a proton = ( ) charge on an electron = 1.6 10 19 C Charge is conserved In an isolated system, total charge does not change with time, though individual charge may change i.e. charge can neither be created nor destroyed. Conservation of charge is also found to hold good in all types of reactions either chemical (atomic) or nuclear. No exceptions to the rule have ever been found. PHYSICS WALLAH 1 ELECTROSTATICS Charge is invariant Charge is independent of frame of reference. i.e. charge on a body does not change whatever be its speed. Attraction Repulsion Similar charges repel each other while dissimilar attract METHODS OF CHARGING Friction : If we rub one body with other body, electrons are transferred from one body to the other. Positive charge Negative charge Glass rod Silk cloth Woollen cloth Rubber shoes, Amber, Plastic objects Dry hair Comb Flannel or cat skin Ebonite rod Note : Clouds become charged by friction Electrostatic induction If a charged body is brought near a metallic neutral body, the charged body will attract opposite charge and repel similar charge present in the neutral body. As a result of this one side of the neutral body becomes negative while the other positive, this process is called 'electrostatic induction'. Charging a body by induction (in four successive steps) charging body q'=0 charged body is brought near uncharged body step-1 charging body q'=-ve uncharged body is connected to earth step-2 charging body q'=-ve uncharged body is disconnected from the earth step-3 q'=-ve charging body is removed step-4 Some important facts associated with induction(i) Inducing body neither gains nor loses charge (ii) The nature of induced charge is always opposite to that of inducing charge (iii) Induction takes place only in bodies (either conducting or non conducting) and not in particles. Conduction The process of transfer of charge by contact of two bodies is known as conduction. If a charged body is put in contact with uncharged body, the uncharged body becomes charged due to transfer of electrons from one body to the other. The charged body loses some of its charge (which is equal to the charge gained by the uncharged body) The charge gained by the uncharged body is always lesser than initial charge present on the charged body. PHYSICS WALLAH 2 ELECTROSTATICS Flow of charge depends upon the potential difference of both bodies. [No potential difference No conduction]. Positive charge flows from higher potential to lower potential, while negative charge flows from lower to higher potential. GOLDEN KEY POINTS Charge differs from mass in the following sense. (i) In SI units, charge is a derived physical quantity while mass is fundamental quantity. (ii) Charge is always conserved but mass is not (Note : Mass can be converted into energy E = mc2 (iii) The quanta of charge is electronic charge while that of mass it is yet not clear. (iv) For a moving charged body mass increases while charge remains constant. True test of electrification is repulsion and not attraction as attraction may also take place between a charged and an uncharged body and also between two similarly charged bodies. For a non relativistic (i.e. v << c) charged particle, specific charge q m = constant Example When a piece of polythene is rubbed with wool, a charge of 2 10 7 C is developed on polythene. What is the amount of mass, which is transferred to polythene. Solution From Q = ne, So, the number of electrons transferred n = Q e = 2 10 7 1.6 10 19 = 1.25 1012 Now mass of transferred electrons = n mass of one electron = 1.25 1012 9.1 10 31 = 11.38 10 19 kg Example 1012 particles (Nuclei of helium) per second falls on a neutral sphere, calculate time in which sphere gets charged by 2 C. Solution Number of particles falling in t second = 1012t Charge on particle = +2e, So charge incident in time t = (1012t).(2e) Given charge is 2 C 2 10 = (10 t).(2e) t = 6 12 10 18 1.6 10 19 = 6.25 s COULOMB'S LAW The electrostatic force of interaction between two static point electric charges is directly proportional to the product of the charges, inversely proportional to the square of the distance between them and acts along the straight line joining the two charges. If two points charges q1 and q2 separated by a distance r. Let F be the electrostatic force between these two charges. According to Coulomb's law. F q1 q2 and F Fe = kq1q 2 r2 where k = PHYSICS WALLAH + 1 F2 on1 q1 r2 1 4 0 = 9 10 9 Nm 2 C2 + q2 F1 on 2 q1 F2 on1 + F1 on 2 q2 , coulomb's constant or electrostatic force constant 3 ELECTROSTATICS Coulomb's law in vector form F12 = force on q1 due to q2 = F21 kq1q 2 = kq 1q 2 F12 r2 r 12 (here r 12 is unit vector from q1 to q2) r2 F21 r q1 q2 r12 Coulomb's law in terms of position vector F12 = kq1q 2 r1 r2 3 y (r1 r2 ) q2 r21 q1 r1 r2 x O Principle of superposition The force is a two body interaction, i.e., electrical force between two point charges is independent of presence or absence of other charges and so the principle of superposition is valid, i.e., force on a charged particle due to number of point charges is the resultant of forces due to individual point charges, i.e., F1 = F12 + F13 +... Note : Nuclear force is many body interaction, so principle of superposition is not valid in case of nuclear force. When a number of charges are interacting, the total force on a given charge is vector sum of the forces exerted on it by all other charges individually F= kq 0 q1 2 1 r r 10 + kq 0 q 2 2 2 r r 20 + ... + kq 0 q i 2 i r r i0 + ... kq 0 q n 2 n r r n 0 in vector form F = kq 0 n i =1 qi ri2 r i0 SOME IMPORTANT POINTS REGARDING COULOMB S LAW AND ELECTRIC FORCE The force is conservative, i.e., work done in moving a point charge once round a closed path under the action of Coulomb s force is zero. The net Coulomb s force on two charged particles in free space and in a medium filled upto infinity are F= 1 4 0 q 1q 2 r 2 and F' = 1 4 q 1q 2 r 2 . So F F' = 0 = K, Dielectric constant (K) of a medium is numerically equal to the ratio of the force on two point charges in free space to that in the medium filled upto infinity. The law expresses the force between two point charges at rest. This law is not applicable for bodies of finite size Although net electric force on both particles change in the presence of dielectric but force due to one charge particle on another charge particle does not depend on the medium between them. Electric force between two charges does not depend on neighboring charges. PHYSICS WALLAH 4 ELECTROSTATICS Example If the distance between two equal point charges is doubled and their individual charges are also doubled, what would happen to the force between them? Solution F= 1 q q 4 0 r2 ....(i) Again, F' = 1 (2q) (2q) 4 0 (2r) 2 4q 2 1 or F' = 4 0 4r 2 = 1 q2 4 0 r 2 =F So, the force will remain the same. Example A particle of mass m carrying charge '+q1' is revolving around a fixed charge ' q2' in a circular path of radius r. Calculate the period of revolution. Solution 1 q 1q 2 4 0 r2 2 T = 2 = mr = (4 0 )r 2 (4 2 mr) q 1q 2 4 2 mr T2 or T = 4 r 0 mr q 1q 2 Example The force of repulsion between two point charges is F, when these are at a distance of 1 m. Now the point charges are replaced by conducting spheres of radii 25 cm having the charge same as that of point charges. The distance between their centres is 1 m, then compare the force of repulsion in two cases. Solution In 2nd case due to mutual repulsion, the effective distance between their centre of charges will be increased (d' > d) so force of repulsion decreases as F 1 d 2 d d' EQUILIBRIUM OF CHARGED PARTICLES In equilibrium net electric force on every charged particle is zero. The equilibrium of a charged particle, under the action of Colombian forces alone can never be stable. Equilibrium of three point charges x Q1 q r Q2 KQ1q (i) Two charges must be of like nature as Fq = (ii) Third charge should be of unlike nature as FQ = PHYSICS WALLAH x 1 2 + KQ 2 q ( r x )2 KQ1Q 2 r 2 + =0 KQ1q x2 =0 5 ELECTROSTATICS Equilibrium of symmetric geometrical point charged system Value of Q at centre for which system to be in state of equilibrium q a q a a a Q q q a (i) For equilateral triangle Q = q a Q q q a (ii) For square Q = 3 q q(2 2 + 1) 4 Equilibrium of suspended point charge system For equilibrium position Tcos Fe Q T Q Tsin x mg Tcos = mg and tan = Fe mg = T sin = Fe = kQ 2 x2 kQ 2 x 2 mg If whole set up is taken into an artificial satellite (geff 0) then T = Fe= kq 4 2 2 2 q q 180 Example For the system shown in figure find Q for which resultant force on q is zero. Solution For force on q to be zero, charges q and Q must be of opposite of nature. Net attraction force on q due to charges Q = Repulsion force on q due to q q (0,a) a a (0,0) Q a FA 2 FA = FR 2 kQq a2 = Q (a,a) FA q FR kq 2 ( 2a ) 2 q = 2 2 Q Hence q = 2 2 Q PHYSICS WALLAH 6 ELECTROSTATICS Example Two identically charged spheres are suspended by strings of equal length. The strings make an angle of 30 with each other. When suspended in a liquid of density 0.8 g/cc the angle remains same. What is the dielectric constant of liquid. Density of sphere = 1.6 g/cc. Solution When set up shown in figure is in air, we have tan 15 = F mg 30 0 15 0 F mg When set up is immersed in the medium as shown in figure, the electric force experienced by the ball will reduce and will be equal to F and the effective r s gravitational force will become mg 1 Thus we have tan F 15 = s mg r 1 = 0 30 F 15 mg F r mgeff=mg 1 1 r = =2 1 s Example Given a cube with point charges q on each of its vertices. Calculate the force exerted on any of the charges due to rest of the 7 charges. Solution Z The net force on particle A can be given by vector sum of force (a,0,a) 1 experienced by this particle due to all the other charges on 4 (a,a,a) (0,a,a) vertices of the cube. For this we use vector form of coulomb's 3 2 law F = kq1q 2 r1 r2 3 ( r1 r2 ) (a,0,0) From the figure the different forces acting on A are given as kq 2 FA1 = ( ) (a,a,0) 7 y a3 ( ) , F = kq ( ai aj ak ) ; F = kq ( ai ak ) = ( 2a ) ( 3a ) ( 2a ) kq ( ai ) kq ( aj ) kq ( ai aj ) , F = , F = = a a 2a ( ) 2 2 A3 3 2 FA 5 (0,a,0) 6 ak kq 2 aj ak FA 2 A (0,0,0) 5 X A4 3 2 2 3 A6 3 3 A7 3 The net force experienced by A can be given as Fnet = FA1 + FA 2 + FA3 + FA 4 + FA5 + FA 6 + FA 7 = PHYSICS WALLAH 1 kq 2 1 + + 1 i + j + k 2 a 2 3 3 ( ) 7 ELECTROSTATICS Example Five point charges, each of value +q are placed on five vertices of a regular hexagon of side Lm. What is the magnitude of the force on a point charge of value q coulomb placed at the centre of the hexagon? Solution If there had been a sixth charge +q at the remaining vertex of hexagon force due to all the six charges on q at O will be zero (as the forces due to individual charges will balance each other). Now if f is the force due to sixth charge and F due to remaining five charges. F + f = 0 F = f F = f = 1 q q 4 0 L2 q2 = 4 0 L2 Do yourself 1 : (i) Find force on charge q placed at the centre of the equilateral triangle as shown. 2Q a a q Q (ii) a Q Two charges Q and 4Q are placed at a separation d, Find the position of a third charge q between them such that it does not experience any force. ELECTRIC FIELD In order to explain action at a distance , i.e., force without contact between charges it is assumed that a charge or charge distribution produces a field in space surrounding it. So the region surrounding a charge (or charge distribution) in which its electrical effects are perceptible is called the electric field of the given charge. Electric field at a point is characterized either by a vector function of position E called electric intensity or by a scalar function of position V called electric potential . The electric field in a certain space is also visualized graphically in terms of lines of force. So electric intensity, potential and lines of force are different ways of describing the same field. Intensity of electric field due to point charge Electric field intensity is defined as force on unit test charge. p r E = Lim q0 0 F q0 = kq r 2 r = kq r 3 r q Note : Test charge (q0) is a fictitious charge and its value is kept small so that it does not change the original charge configuration. PHYSICS WALLAH 8 ELECTROSTATICS Properties of electric field intensity : (i) It is a vector quantity. Its direction is the same as the force experienced by positive test charge. (ii) Electric field due to positive charge is always away from it while due to negative charge always towards it. (iii) Its unit is Newton/coulomb (iv) Its dimensional formula is [MLT 3A 1] (v) Force on a point charge is in the same direction of electric field on positive charge and in opposite direction on a negative charge. F = qE (vi) It obeys the superposition principle that is the field intensity point due to charge distribution is vector sum of the field intensities due to individual charges GOLDEN KEY POINTS Charged particle in an electric field always experiences a force either it is at rest or in motion. In presence of a dielectric, electric field decreases and becomes Test charge is always a positive charge. r times of its value in free space. Ftest = E 1 test charge If identical charges are placed on each vertices of a regular polygon, then E at centre = zero. ELECTRIC FIELD INTENSITIES DUE TO VARIOUS CHARGE DISTRIBUTIONS Due to discrete distribution of charge p th By principle of superposition intensity of electric field due to i charge E ip = kq ri 3 ri q4 q1 ri q3 Net electric field due to whole distribution of charge E p = E i qi q2 i =1 Continuous distribution of charge r Treating a small element as particle E = Due to linear charge distribution E= Due to surface charge distribution E = k 1 4 0 k s Due to volume charge distribution E = k v PHYSICS WALLAH dq d r 2 ds r2 dv r2 r 3 r P dq [ = charge per unit length] [ = charge per unit area] [ = charge per unit volume] 9 ELECTROSTATICS Electric field strength at a general point due to a uniformly charged rod As shown in figure, if P is any general point in the surrounding of rod, to find electric field strength at P, we consider an element on rod of length dx at a distance x from point O as shown in figure. Now if dE be the electric field at P due to the P element, then dE = (x kdq 2 + r2 ) Q Here dq = L 1 2 dx r kQ sin kdq dEx = dE sin = 2 2 sin = dx x +r ) L ( x2 + r2 ) ( Here we have x = r tan and dx = r sec2 d kQ r sec 2 d kQ sin = sin d 2 2 r sec L Lr Thus dEx = Ex = O L dE x = kQ Lr + 1 2 sin d = kQ Lr cos + 1 2 = kQ Lr [cos 2 cos 1] Similarly, electric field strength at point P due to dq in y direction is dEcos kQdx dEy = dEcos = L (r2 + x2 ) Again we have x = r tan and dx = r sec d . Thus we have 2 dEy= Ey = = kQ L dE y = kQ Lr cos kQ Lr r sec 2 r 2 sec 2 d = kQ Lr cos d 2 = kQ Lr dEsin P r cos d dx + 1 dE cos + sin x + 1 2 [sin 1 + sin 2] Thus electric field at a general point in the surrounding of a uniformly charged rod which subtend angles 1 and 2 at the two corners of rod can be given as in x direction : Ex = kQ Lr (cos 1 cos 2) and in y direction E y = kQ Lr (sin 1 + sin 2) Electric field due to a uniformly charged ring Case I : At its centre B A+ + + + + + Here by symmetry we can say that electric field strength at centre due + + + + + + O R + + centre due to the segment exactly opposite to it. The electric field + to every small segment on ring is cancelled by the electric field at + + strength at centre due to segment AB is cancelled by that due to + + + + + + PHYSICS WALLAH + uniformly charged ring is E0 = 0 + segment CD. This net electric field strength at the centre of a D C 10 ELECTROSTATICS Case II : At a point on the axis of Ring Here we'll find the electric field strength at point P due to the ring which is situated at a distance x from the ring centre. For this we consider a small section of length d on ring as Q shown. The charge on this elemental section is dq = 2 R d [Q= total charge of ring] d + + + + + + + + + R + C + + P dE cos dE sin dE + + + + + Q Due to the element d , electric field strength dE at point P can be given as dE = (R Kdq 2 + x2 ) The component of this field strength dE sin which is normal to the axis of ring will be cancelled out due to the ring section opposite to d . The component of electric field strength along the axis of ring dE cos due to all the sections will be added up. Hence total electric field strength at point P due to the ring is Ep = dE cos = ( R = kdq 2 +x 2 ) x R2 + x2 kQx 2 R ( R 2 + x 2 ) = 3/ 2 2 R = kQx 2 R ( R 2 +x 2 ) d = kQx 2 R ( R 2 + x 2 ) 3/ 2 d kQx (R 2 + x2 ) 3/ 2 Electric field strength due to a charged circular arc at its centre : Figure shows a circular arc of radius R which subtend an angle at its centre. To find electric field strength at C, we consider a polar segment on arc of angular width d at an angle from the angular bisector XY as shown. Rd +++++++++ + + + ++ ++ + ++ d ++ + ++ ++ + x ++ R C dEsin dE dEcos Y Q .d The length of elemental segment is Rd , the charge on this element d is dq = Due to this dq, electric field at centre of arc C is given as dE = PHYSICS WALLAH kdq R2 11 ELECTROSTATICS Now electric field component due to this segment dEsin which is perpendicular to the angular bisector gets cancelled out in integration and net electric field at centre will be along angular bisector which can be calculated by integrating dEcos within limits from 2 to 2 . Hence net electric field strength at centre C is E C = dE cos + / 2 = / 2 kQ R 2 cos d = + / 2 kQ R 2 cos d = / 2 kQ R 2 sin / 2 + / 2 2kQ sin kQ 2 sin + sin = = 2 2 R 2 2 R Electric field strength due to a uniformly surface charged disc : If there is a disc of radius R, charged on its surface with surface charge density we wish to find electric field strength due to this disc at a distance x from the centre of disc on its axis at point P shown in figure. dy y x P dE To find electric field at point P due to this disc, we consider an elemental ring of radius y and width dy in the disc as shown in figure. The charge on this elemental ring dq = .2 ydy [Area of elemental ring ds = 2 y dy] Now we know that electric field strength due to a ring of radius R, charge Q, at a distance x from its centre E= on its axis can be given as kQx (x 2 + R2 ) 3/ 2 Here due to the elemental ring electric field strength dE at point P can be given as dE = kdqx (x 2 +y 2 ) 3/ 2 = k 2 y dy x (x 2 + y2 ) 3/ 2 Net electric field at point P due to this disc is given by integrating above expression from 0 to R as E = dE = k 2 xydy R 0 (x = 2k x PHYSICS WALLAH 2 + y2 ) 3/ 2 R = k x 0 = 2 0 x 2 + y2 0 1 R 2y dy (x 2 1 + y2 ) 3/ 2 x +R x 2 2 12 ELECTROSTATICS Example Calculate the electric field at origin due to infinite number of charges as shown in figures below. fig (a) O fig (b) q q q 1 2 4 x(m) O q q q 1 2 4 x(m) Solution (a) E 0 = kq 4kq a 1 kq.1 1 1 1 1 + 4 + 16 + = (1 1 / 4) = 3 [ S = 1 r , a = 1 and r = 4 ] (b) E 0 = kq kq.1 4 kq 1 1 1 1 4 + 16 = 1 ( 1 / 4 ) = 5 ( ) Example A charged particle is kept in equilibrium in the electric field between the plates of millikan oil drop experiment. If the direction of the electric field between the plates is reversed, then calculate acceleration of the charged particle. Solution Let mass of the particle = m, Charge on particle = q Intensity of electric field in between plates = E, Initially mg = qE After reversing the field ma = mg + qE ma = 2mg Acceleration of particle a = 2g Example Calculate the electric field intensity E which would be just sufficient to balance the weight of an electron. If this electric field is produced by a second electron located below the first one what would be the distance between them? [Given : e = 1.6 10C, m = 9.1 10 31 kg and g= 9.8 m/s2] 19 Solution As force on a charge e in an electric field E Fe = eE So according to given problem Fe = W eE = mg E= mg e = 9.1 10 31 9.8 1.6 10 19 = 5.57 10 11 V m As this intensity E is produced by another electron B, located at a distance r below A. E= So, 1 e 4 0 r 2 r = e 4 0 E 9 109 1.6 10 19 r= 5.57 10 11 PHYSICS WALLAH 5m 13 ELECTROSTATICS Example A block having mass m = 4 kg and charge q = 50 C is connected to a spring having a force constant k = 100 N/m. The block lies on a frictionless horizontal track and a uniform electric field E = 5 105 V/m acts on the system as shown in figure. The block is released from rest when the spring is unscratched (at x = 0) (a) By what maximum amount does the spring expand? (b) What is the equilibrium position of the block? (c) Show that the block's motion is simple harmonic and determine the amplitude and time period of the motion. Solution (a) As x increases, electric force qE will accelerate the block while elastic force in the spring kx will oppose the motion. The block will move away from its initial position x = 0 till it comes to rest, i.e., work done by the electric force is equal to the energy stored in the spring. So if xmax is maximum stretch of the spring. 1 2 (b) kx 2max = (qE)x max xmax = k xmax = 2 (50 10 6 ) (5 10 5 ) 100 = 0.5 m In equilibrium position FR = 0, so if x0 is the stretch of the spring in equilibrium position kx0 = qE x0 = (qE/k) = (c) 2qE 1 2 xmax = 0.25 m If the displacement of the block from equilibrium position (x0) is x, restoring force will be F = k(x x0) qE = kx [ax kx0 = qE] and as the restoring force is linear the motion will be simple harmonic with time period T = 2 m k = 2 4 100 = 0.4 s and amplitude = xmax x0 = 0.5 0.25 = 0.25 m Do yourself 2 : (i) Find Electric field at the centre C of the square as shown : (ii) Three equal charges (Q) are placed on three vertices of a square of side 'a' as shown. Find the magnitude of electric field at the remaining vertex. PHYSICS WALLAH 14 ELECTROSTATICS ELECTRIC LINES OF FORCE Electric lines of electrostatic field have following properties + A B (i) Imaginary (ii) Can never cross each other (iii) Can never be closed loops (iv) The number of lines originating or terminating on a charge is proportional to the magnitude of qA>qB charge. In rationalised MKS system (1/ o) electric lines are associated with unit charge, so if a body encloses a charge q , total lines of force associated with it (called flux) will be q/ o. (v) Lines of force ends or starts normally at the surface of a conductor. (vi) If there is no electric field there will be no lines of force. (vii) Lines of force per unit area normal to the area at a point represents magnitude of intensity, crowded lines represent strong field while distant lines weak field. (viii) Tangent to the line of force at a point in an electric field gives the direction of intensity. So a positive charge free to move follow the line of force. GOLDEN KEY POINTS Lines of force starts from (+ve) charge and ends on ( ve) charge. Lines of force start and end normally on the surface of a conductor . + fixed point charge near infinite metal plate +s + + E=0 + + + s E=0 edge effect The lines of force never intersect each other due to superposition principle. The property that electric lines of force contract longitudinally leads to explain attraction between opposite charges. The property that electric lines of force exert lateral pressure on each other leads to explain repulsion between like charges. Electric flux ( ) The word "flux" comes from a Latin word meaning "to flow" and you can consider the flux of a vector field to be a measure of the flow through an imaginary fixed element of surface in the field. Electric flux is defined as E = E dA This surface integral indicates that the surface in question is to be divided into infinitesimal elements of area dA and the scalar quantity E dA is to be evaluated for each element and summed over the entire surface. PHYSICS WALLAH 15 ELECTROSTATICS Important points about electric flux : (i) It is a scalar quantity (ii) Units (V m) and N m2/C Dimensions : [ML3T 3A 1] (iii) The value of does not depend upon the distribution of charges and the distance between them inside the closed surface. Electric Flux through a circular Disc : Figure shows a point charge q placed at a distance from a disc of radius R. Here we wish to find the electric flux through the disc surface due to the point charge q. We know a point charge q originates electric flux in radially outward direction. The flux is originated in cone shown in figure passes through the disc surface. q To calculate this flux, we consider on elemental ring an disc surface of radius x and width dx as shown. Area of this ring (strip) is dS = 2 x dx. The electric field due to q at this elemental ring is given as E= (x kq 2 + 2 ) If d is the flux passing through this elemental ring, then dS x = R d = (x q 2 0 = 2 0 kq 2 + 2 d = EdS cos = q ) 2 x dx x dx 0 ( 2 + x2 ) 3/ 2 = q 2 0 x + 2 R 0 q 1 = + x 2 0 2 0 ( 1 2 = 2 kq x dx ( 2 + x2 ) 3/ 2 x dx 2 R q 2 + x2 ) 3/ 2 + x2 1 2 The above result can be obtained in a much simpler way by using the concept of solid angle and Gauss's law. PHYSICS WALLAH 16 ELECTROSTATICS Electric flux through the lateral surface of a cylinder due to a point charge : Figure shows a cylindrical surface of length L and radius R. On its axis at its centre a point charge q is placed. Here we wish to find the flux coming out from the lateral surface of this cylinder due to the point charge q. For this we consider an elemental strip of width dx on the surface of cylinder as shown. The area of this strip is dS = 2 R.dx dS E x C q R L The electric field due to the point charge on the strip can be given as E = (x kq 2 + R2 ) . If d is the electric flux through the strip, then d = ES cos = (x Kq 2 +R 2 ) 2 Rdx R x +R 2 = 2 KqR2 2 dx (x 2 + R2 ) 3/ 2 Total flux through the lateral surface of cylinder = d = qR 2 2 0 +L/2 L/2 dx (x 2 + R2 ) 3/ 2 = q 0 2 + 4R 2 This situation can also be easily handled by using the concepts of Gauss's law. GAUSS'S LAW It relates with the total flux of an electric field through a closed surface to the net charge enclosed by that surface and according to it, the total flux linked with a closed surface is (1/ 0) times the charge enclosed by the closed surface i.e., S E.ds = q 0 E n ds r O +q r Regarding Gauss's law it is worth noting that : (i) Flux through gaussian surface is independent of its shape. (ii) (iii) (iv) (v) Flux through gaussian surface depends only on charges present inside gaussian surface. Flux through gaussian surface is independent of position of charges inside gaussian surface. Electric field intensity at the gaussian surface is due to all the charges present (inside as well as out side) In a close surface incoming flux is taken negative while outgoing flux is taken positive. (vi) In a gaussian surface = 0 does not employ E = 0 but E = 0 employs = 0. PHYSICS WALLAH 17 ELECTROSTATICS (vii) Gauss's law and Coulomb's law are equivalent, i.e., if we assume Coulomb's law we can prove Gauss's law and vice versa. To prove Gauss's law from Coulomb's law consider a hypothetical spherical surface [called Gaussian surface] of radius r with point charge q at its centre as shown in figure. By Coulomb's law intensity at a point P on the surface will be, E = 1 4 0 r 3 r And hence electric flux linked with area ds E.ds = 1 q 4 0 r 3 r.ds Here direction of r and ds are same, i.e., E .ds = S 1 q 4 0 r 2 dS S = 1 1 4 0 r 2 ( 4 r ) 2 q E.ds = S (viii) 0 Which is the required result. Though here in proving it we have assumed the surface to be spherical, it is true for any arbitrary surface provided the surface is closed. (a) If a closed body (not enclosing any charge) is placed in an electric field (either uniform or non uniform) total flux linked with it will be zero. sphere E=0 (B) E=0 (A) (b) If a closed body encloses a charge q, then total flux q (A) (B) E=(q/ 0) (C) q E=(q/ 0) q (D) q E=(q/ 0) E=(q/ 0) linked with the body will be S E.ds = q 0 From this expression it is clear that the flux linked with a closed body is independent of the shape and size of the body and position of charge inside it.[figure] Note : So in case of closed symmetrical body with charge at its centre, flux linked with each half will be 1 2 ( E ) = q and the symmetrical closed body has n identical faces with point 2 0 q E = n n 0 charge at its centre, flux linked with each face will be PHYSICS WALLAH 18 ELECTROSTATICS (ix) Gauss's law is a powerful tool for calculating electric intensity in case of symmetrical charge distribution by choosing a Gaussian surface in such a way that E is either parallel or perpendicular to its various faces. As an example, consider the case of a plane sheet of charge having charge density . To calculate E at a point P close to it consider a Gaussian surface in the form of a 'pill box' of cross section S as shown in figure. + E n + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + E n n n E n Ein=0 Conductor (B) Sheet of charge (A) The charge enclosed by the Gaussian surface = S and the flux linked with the pill box = ES + 0 + ES = 2ES (as E is parallel to curved surface and perpendicular to plane faces) So from Gauss's law, E = (x) If E = 0, = E.ds 1 0 (q), 2ES = 1 0 ( S) E = = 0, so q = 0 but if q = 0, 2 0 E.ds = 0 So E may or may not be zero. E.ds If a dipole is enclosed by a closed surface then, q =0, so =0, but E 0 Note : If instead of plane sheet of charge, we have a charged conductor, then as shown in figure (B) Ein= 0. So E = ES and hence in this case E = 0 . This result can be verified from the fact that intensity at the surface of a charged spherical conductor of radius R is, E = 1 q 4 0 R 2 with q = 4 R2 So for a point close to the surface of conductor, E = 1 4 0 R 2 ( 4 R 2 ) = 0 Example If a point charge q is placed at the centre of a cube. What is the flux linked (a) with the cube? (b) with each face of the cube? Solution (a) According to Gauss's law flux linked with a closed body is (1/ 0) times the charge enclosed and here the closed body cube is enclosing a charge q so, T = (b) 1 0 (q) Now as cube is a symmetrical body with 6 faces and the point charge is at its centre, so electric flux linked with each face will be F = Note: (i) (ii) 1 6 ( T ) = q 6 0 Here flux linked with cube or one of its faces is independent of the side of cube. If charge is not at the centre of cube (but anywhere inside it), total flux will not change, but the flux linked with different faces will be different. PHYSICS WALLAH 19 ELECTROSTATICS Example If a point charge q is placed at one corner of a cube, what is the flux linked with the cube? Solution In this case by placing three cubes at three sides of given cube and four cubes above, the charge will be in the centre. So, the flux linked with each cube will be one eight of the flux q . Flux associated with given cube = 0 q 8 0 FLUX CALCULATION USING GAUSS LAW ds E E in= R2 E and out= R2 E R ds ds E total = 0 E R in= circular= R2 E and out= curved = R2 E total =0 E y ds E a ds in= a2 E and out= a2 E total = 0 ds x a a z E in = 1 2 R E and 2 out = 1 2 R E total = 0 2 E E= R q kq R q 2 = 2 R 2 4 0 R 2 = q 2 0 Note : here electric field is radial hemisphere= q q q 2 0 q cylinder = q R q 2 0 R PHYSICS WALLAH 20 ELECTROSTATICS q q cube = q q = q 2 0 q 8 0 q q q q = q 4 0 Example As shown in figure a closed surface intersects a spherical conductor. If a negative charge is placed at point P. What is the nature of the electric flux coming out of the closed surface ? conductor ++++ close surface + ++ ++++ P Q + ++ + Solution Point charge Q induces charge on conductor as shown in figure. Net charge enclosed by closed surface is negative so flux is negative. Example Consider E = 3 103 i (N/C) then what is the flux through the square of 10 cm side, if the normal of its plane makes 60 angle with the X axis. Solution = EScos = 3 103 [10 10 2]2 cos60 = 3 103 10 2 = 15 Nm2/C Example Find the electric field due to an infinitely long cylindrical charge distribution of radius R and having linear charge density at a distance half of the radius from its axis. Solution r= R 2 point will be inside PHYSICS WALLAH so E= 2k r R 2 = 2k R R 2 = 4 0 R 2 21 ELECTROSTATICS Do yourself 3 : (i) (ii) (iii) If a point charge q is placed at the centre of a face of a cube. Find the flux through the remaining faces of the cube. Find the flux through a spherical surface of radius R in a uniform Electric field E. Find the flux of Electric field through the spherical surface, where an infinitely long line charge (linear charge density ) passes along the diameter of the sphere as shown: + + + + R + + + + ELECTRIC FIELD DUE TO SOLID CONDUCTING OR HOLLOW SPHERE For outside point (r > R) + r + 0 + + ds E .ds = q q is Ep = 0 + + Gaussian surface [E 0 2 E 4 r = + + O R + At every point on the Gaussian surface E ds ; E .ds = E ds cos 0 = E dA P + q + Using Gauss's theorem E.d s = E + constant over the gaussian surface] q 4 0 r 2 q For surface point r = R : For Inside point (r < R) : Because charge inside the conducting sphere or hollow is zero. ES = 4 0 R (i.e. q = 0) So 2 q E .ds = = 0 Ein = 0 0 ELECTRIC FIELD DUE TO SOLID NON CONDUCTING SPHERE Outside (r > R) From Gauss's theorem q E .ds = E 4 r 2 = 0 EP = P q 0 E ds r O R q 4 0 r 2 Gaussian surface At surface (r = R) ES = q 4 0 R 2 PHYSICS WALLAH Put r = R 22 ELECTROSTATICS Inside (r < R) : E From Gauss's theorem q E .ds = P 0 s ds r O R Where q charge contained within Gaussian surface of radius r E(4 r 2 ) = q 0 E= q Gaussian surface ...(i) 4 r 2 0 As the sphere is uniformly charged, the volume charge density (charge/volume) is constant throughout the sphere = q charge enclosed in gaussian surface 4 3 R 3 q 4 q = r 3 = 3 3 (4 3) R 4 3 r 3 put this value in equation (i) Ein = 1 q = qr 3 R3 qr 4 0 R 3 ELECTRIC FIELD DUE TO AN INFINITE LINE DISTRIBUTION OF CHARGE Let a wire of infinite length is uniformly charged having a constant linear charge density . P is the point where electric field is to be calculated. dS1 Let us draw a coaxial Gaussian cylindrical surfaces of length . r E From Gauss's theorem E .dS + E .dS + E .dS 1 2 s1 s2 E dS1 3 = s3 so E .dS1 = 0 and E 2 r = q 0 q 0 E dS 2 so dS3 E E .dS 2 = 0 E dS2 [ E dS 3 ] zd Gaussian surface Charge enclosed in the Gaussian surface q = . So or E 2 r = E= 2k 0 E= where k = r (f) 2 0 r 1 4 0 Gaussian Surface Charged cylindrical nonconductor of infinite length Electric field at outside point Electric field at inside point PHYSICS WALLAH EA = EB = 2k r r>R r r 2 0 R 2 r<R 23 ELECTROSTATICS DIELECTRIC IN ELECTRIC FIELD Let E 0 be the applied field, Due to polarisation, electric field is E p . E0 The resultant field is E . For homogeneous and isotropic dielectric, the direction of E p is opposite to the direction of E 0 . E0 Ep E + E 0 + + + + So, Resultant field is E = E0 EP GOLDEN KEY POINTS Electric field inside a solid conductor is always zero. Electric field inside a hollow conductor may or may not be zero (E 0 if non zero charge is inside the sphere). The electric field due to a circular loop of charge and a point charge are identical provided the distance of the observation point from the circular loop is quite large as compared to its radius i.e. x >>> R. Example For infinite line distribution of charge draw the curve between log E and log r. log E Solution E= 2 0 r = A where A = r 2 0 = constant log A take log on both side log E = log A log r log r Example A point charge of 0.009 C is placed at origin. Calculate intensity of electric field due to this point charge at point ( 2, 7, 0 ) . Solution E= qr 4 0 r 3 ; where r = xi + yj = 2i + 7 j , E= 9 10 5 9 10 9 ( 2 i + 7 j ) (3) 3 = ( 3 2 i + 3 7 j ) NC 1 ELECTROSTATIC POTENTIAL ENERGY Potential energy of a system of particles is defined only in conservative fields. As electric field is also conservative, we define potential energy in it. Potential energy of a system of particles we define as the work done in assembling the system in a given configuration against the interaction forces of particles. Electrostatic potential energy is defined in two ways. (i) Interaction energy of charged particles of a system (ii) Self energy of a charged object PHYSICS WALLAH 24 ELECTROSTATICS Electrostatic Interaction Energy Electrostatic interaction energy of a system of charged particles is defined as the external work required to assemble the particles from infinity to the given configuration. When some charged particles are at infinite separation, their potential energy is taken zero as no interaction is there between them. When these charges are brought close to a given configuration, external work is required if the force between these particles is repulsive and energy is supplied to the system, hence final potential energy of system will be positive. If the force between the particle is attractive, work will be done by the system and final potential energy of system will be negative. Interaction Energy of a system of two charged particles Figure shows two + ve charges q1 and q2 separated by a distance r. The electrostatic interaction energy of this system can be given as work done in bringing q2 from infinity to the given separation from q1. x r q1 q2 r It can be calculated as W = F .dx = W= kq 1q 2 r r kq1q 2 x2 dx F dx [ ve sign shows that x is decreasing] = U [interaction energy] If the two charges here are of opposite sign, the potential energy will be negative as U = kq 1q 2 r Interaction Energy for a system of charged particles When more than two charged particles are there in a system, the interaction energy can be given by sum of interaction energies of all the pairs of particles. For example if a system of three particles having charges q1, q2 and q3 is given as shown in figure. q1 r3 r2 r1 q2 q3 The total interaction energy of this system can be given as U = kq 1q 2 r3 + kq 1q 3 r2 + kq 2 q 3 r1 ELECTRIC POTENTIAL Electric potential is a scalar property of every point in the region of electric field. At a point in electric field potential is defined as the interaction energy of a unit positive charge. If at a point in electric field a charge q0 has potential energy U, then electric potential at that point can be given as V = U q0 joule/coulomb Potential energy of a charge in electric field is defined as work done in bringing the charge from infinity to the given point in electric field. Similarly we can define electric potential as "work done in bringing a unit positive charge from infinity to the given point against the electric forces. PHYSICS WALLAH 25 ELECTROSTATICS Example A charge 2 C is taken from infinity to a point in an electric field, without changing its velocity, if work done against electrostatic forces is 40 J then potential at that point is ? Solution V= Wext q Note : = 40 J 2 C = 20V Always remember to put sign of W and q. Electric Potential due to a point charge in its surrounding : The potential at a point P at a distance r from the charge q VP = kqq 0 U is the potential energy of charge q0 at point p, U = point P is VP = r U q0 p . Where r q . Thus potential at kq r Electric Potential due to a charge Rod : Figure shows a rod of length L, uniformly charged with a charge Q. Due to this we'll find electric potential at a point P at a distance r from one end of the rod as shown in figure. x Q dx r P L For this we consider an element of width dx at a distance x from the point P. Charge on this element is dQ = Q L dx The potential dV due to this element at point P can be given by using the result of a point charge as dV = kdq x = kQ Lx dx : V = dV = Net electric potential at point P r+L r Lx dx = kQ L r+L r ln Electric potential due to a charged ring Case I : At its centre To find potential at the centre C of the ring, we first find potential dV at centre due to an elemental charge dq on ring which is given as dV = + + + R C + + + R = kQ R . + kdq + + V = dV = + dq Total potential at C is + + R Q + + + kdq + + kQ As all dq's of the ring are situated at same distance R from the ring centre C, simply the potential due to all dq's is added as being a scalar quantity, we can directly say that the total electric potential at ring centre is kQ R . Here we can also state that even if charge Q is non uniformly distributed on ring, the electric potential C will remain same. PHYSICS WALLAH 26 ELECTROSTATICS Case II : At a point on axis of ring We find the electric potential at a point P on the axis of ring as shown, we can directly state x 2 + R 2 from the point P, thus the result as here also all points of ring are at same distance kQ the potential at P can be given as VP = R2 + x2 + + + + x 2 +R 2 + + + + R P x + + + + + Electric potential due to a uniformly charged disc : Figure shows a uniformly disc of radius R with surface charge density coul/m2. To find electric potential at point P we consider an elemental ring of radius y and width dy, charge on this elemental ring is dq = 2 y dy. Due to this ring, the electric potential at point P can be kdq k. .2 y dy given as dV = = 2 2 x +y x 2 + y2 dy y x P Net electric potential at Point P due to whole disc can be given as V= dV = R 0 2 0 y dy x2 + y2 = x 2 + y2 = x 2 + R 2 x 0 2 0 2 0 R ELECTRIC POTENTIAL DUE TO HOLLOW OR CONDUCTING SPHERE Case - I At outside sphere R r>R P According to definition of electric potential, electric potential at point P r r V = E .dr = q 4 0 r 2 dr Eout q V= ; = 4 0 4 0 r 2 q r 1 r 2 dr = r q 1 = 4 0 r 4 0 r q Case - II At surface R R V = E.dr = q = 4 0 q 4 0 r 2 dr E out = q 1 ; V= dr 2 4 0 r 2 4 0 r q R R 1 q V= 4 0 R r PHYSICS WALLAH 27 ELECTROSTATICS Case - III Inside the surface : Inside the surface E = V = constant so V = dV dr =0 q 4 0 R ELECTRIC POTENTIAL DUE TO SOLID NON CONDUCTING SPHERE Case-I At outside sphere Same as conducting sphere. Case -II At Surface Same as conducting sphere. Case-III Inside the sphere r R V = E1dr + E 2 dr R r R kq kqr V = 2 dr + 3 dr R R r 1 R kq r 2 r V = kq + 3 r R 2 R 1 r2 R2 V = kq + 3 2R 3 R 2R V= r V = E .dr kq 3R 2 r 2 2R 3 Potential Difference Between Two points in electric field Potential difference between two points in electric field can be defined as work done in displacing a unit positive charge from one point to another against the electric forces. A VA B VB If a unit +ve charge is displaced from a point A to B as shown work required can be given as VB VA = B A E.dx If a charge q is shifted from point A to B, work done against electric forces can be given as W = q(VB VA) If in a situation work done by electric forces is asked, we use W = q(VA VB) If VB < VA, then charges must have tendency to move toward B (low potential point) it implies that electric forces carry the charge from high potential to low potential points. Hence we can say that in the direction of electric field always electric potential decreases. Example 1 C charge is shifted from A to B and it is found that work done by external force is 80 J against electrostic forces, find VA VB Solution WAB = q(VB VA) 80 J = 1 C (VB VA) VA VB = 80 V PHYSICS WALLAH 28 ELECTROSTATICS Equipotential surfaces For a given charge distribution, locus of all points having same potential is called 'equipotential surface'. Equipotential surfaces can never cross each other (otherwise potential at a point will have two values which is absurd) Equipotential surfaces are always perpendicular to direction of electric field. If a charge is moved from one point to the other over an equipotential surface then work done WAB = UAB = q (VB VA) = 0 [ VB = VA] Shapes of equipotential surfaces V=V1 V=V2 V=V1 V=V2 V=V2 for uniform electric field equipotential surfaces are parallel plane V=V1 for a point charge, spherical conductor equipotential surfaces are spherical for a line distribution of charge equipotential surfaces are cylinderical The intensity of electric field along an equipotential surface is always zero. Electric Potential Gradient The maximum rate of change of potential at right angles to an equipotential surface in an electric field is defined as potential gradient. E = V = grad V Note : Potential is a scalar quantity but the gradient of potential is a vector quantity V V V i+ j+ k In cartesian co ordinates V = y z x Example If V = 5x + 3y + 15 z then find magnitude of electric field at point (x,y,z). Solution V V V i+ j+ k = ( 5 i + 3 j + x y z E = 15k ) E = 25 + 9 + 15 = 46 = 7 unit Example The four charges q each are placed at the corners of a square of side a. Find the potential energy of one of the charges The electric potential of a point A due to charges B, C and D is V= 1 q 4 0 a + 1 4 0 q 2a + 1 q 4 0 a = 1 q 2+ 4 0 2 a 1 Potential energy of the charge at A is PE = qV = PHYSICS WALLAH 1 q2 2 + . 4 0 2 a 1 29 ELECTROSTATICS Example A proton moves with a speed of 7.45 105 m/s directly towards a free proton originally at rest. Find the distance of closest approach for the two protons. Given : 1 4 0 = 9 109 N m2 C2 ; mp = 1.67 10 27 kg and e = 1.6 10 19 C Solution As here the particle at rest is free to move, when one particle approaches the other, due to electrostatic repulsion other will also start moving and so the velocity of first particle will decrease while of other will increase and at closest approach both will move with same velocity. So if v is the common velocity of each particle at closest approach, by 'conservation of momentum'. mu = mv + mv v = 1 2 1 And by conservation of energy So, r = 4e 2 4 0 mu 2 [as v = u 2 u 2 2 mu = 1 2 2 mv + 1 2 2 mv + 1 e2 4 0 r ] And hence substituting the given data, r = 9 109 4 (1.6 10 19 ) 2 1.67 10 27 (7.45 10 3 ) 2 = 10 12 m Do yourself - 4 : (i) Four point charges q are placed at the corners of a square of side 'a'. Find the potential energy of the system (ii) Two identical particles of mass m and charge q are released at a separation r from each other. Find their velocities, when they become separated by large distance. (iii) If V = 7x2y + 2y z3x, then find electric field at point (x,y,z). ELECTRIC DIPOLE A system of two equal and opposite charges separated by a certain distance is called electric dipole, shown in figure. Every dipole has a characteristic property called dipole moment. It is defined as the product of magnitude of either charge and the separation between the charges, given as p = qd -q d p +q In some molecules, the centres of positive and negative charges do not coincide. This results in the formation of electric dipole. Atom is non polar because in it the centres of positive and negative charges coincide. Polarity can be induced in an atom by the application of electric field. Hence it can be called as induced dipole. Dipole Moment : Dipole moment p = q d (i) (ii) Vector quantity, directed from negative to positive charge Dimension : [LTA], Units : coulomb metre (or C-m) PHYSICS WALLAH +q p q 30 ELECTROSTATICS Example A system has two charges qA = 2.5 10 7 C and qB = 2.5 10 7 C located at points A: (0, 0, 0.15m) and B ; (0, 0, + 0.15 m) respectively. What is the total charge and electric dipole moment of the system? Solution Total charge = 2.5 10 7 2.5 10 7 = 0 Electric diople moment, p = Mangitude of either charge separation between charges = 2.5 10 7 [0.15 + 0.15] C m = 7.5 10 8 C m. The direction of dipole moment is from B to A. Dipole Placed in uniform Electric Field Figure shows a dipole of dipole moment p placed at an angle to the direction of electric field. Here the charges of dipole experience forces qE in opposite direction as shown. Fnet = qE + ( q) E = 0 + qE p qE E Thus we can state that when a dipole is placed in a uniform electric field, net force on the dipole is zero. But as equal and opposite forces act with a separation in their line of action, they produce a couple which tend to align the dipole along the direction of electric field. The torque due to this couple can be given as = Force separation between lines of actions of forces = qE d sin = pE sin = r F = d qE = qd E = p E Work done in Rotation of a Dipole in Electric field When a dipole is placed in an electric field at an angle , the torque on it due to electric field is = pE sin Work done in rotating an electric dipole from 1 to 2 [ uniform field] so W = dW = dW = d d 2 and W 1 2 =W= pE sin d = pE (cos 1 cos 2) 1 e.g. W0 180 = pE [1 ( 1)] = 2 pE W0 90 = pE (1 0) = pE If a dipole is rotated from field direction ( = 0 ) to then W = pE (1 cos ) + + =0 E = 90 + E p E p p = 180 = minimum = 0 = maximum = pE = minimum = 0 W = minimum = 0 W = pE W = maximum = 2pE PHYSICS WALLAH 31 ELECTROSTATICS Electrostatic potential energy : Electrostatic potential energy of a dipole placed in a uniform field is defined as work done in rotating a dipole from a direction perpendicular to the field to the given direction i.e., W90 = pE sin d = pE cos = p.E , E is a conservative field so whatever work is 90 done in rotating a dipole from 1 to 2 is just equal to change in electrostatic potential energy W = U U = pE (cos 1 cos 2) 1 2 2 1 Work done in rotating an electric dipole in an electric field Suppose at any instant, the dipole makes an angle with the electric field. The torque acting on dipole. = qEd = (q 2 sin )E = pE sin The work done in rotating dipole from 1 to 2 B 2 qE W = 2 2 1 1 d = qE +q d q A C E pE sin d W = pE (cos 1 cos 2) = U2 U1 ( U = pE cos ) Force on an electric dipole in Non uniform electric field : If in a non uniform electric field dipole is placed at a point where electric field is E, the interaction energy of dipole at this point U = p.E . Now the force on dipole due to electric field F = U r If dipole is placed in the direction of electric field then F= p dE dx +++++++++++ Example Calculate force on a dipole in the surrounding of a long charged wire as shown in the figure. -q p +q r Solution In the situation shown in figure, the electric field strength due to the wire, at the position of dipole as E = 2k r Thus force on dipole is F = p. dE dr 2k 2 r = p = 2kp r2 Here ve charge of dipole is close to wire hence net force an dipole due to wire will be attractive. PHYSICS WALLAH 32 ELECTROSTATICS ELECTRIC POTENTIAL DUE TO DIPOLE At axial point Electric potential due to +q charge V1 = Electric potential due to q charge V2 = Net electric potential kq P r (r + ) V = V1 + V2 = kq + (r ) kq (r + ) = kq 2 (r 2 2 ) = kp r 2 2 kp r2 At equitorial point V1 = Electric potential of P due to q charge V2 = kq kq x x P kq Electric potential of P due to +q charge Net potential V = V1 + V2 = q O kq If r > > > V = q (r ) x kq x r x x =0 V=0 q O q At general point V= p cos 4 0 r = 2 p.r 4 0 r 3 p = q d electric dipole moment Electric field due to an electric dipole Figure shows an electric dipole placed on x axis at origin. Here we wish to find the electric field and potential at a point O having coordinates (r, ). Due to the positive charge of dipole electric field at O is in radially outward direction and due to the negative charge it is radially inward as shown in figure. Er = V r = 2kp cos r 3 and E = 1 V r Enet E Er O kp sin = r r3 Thus net electric field at point O, -q kp Enet = E 2r + E 2 = y r x +q 1 + 3cos 2 3 If the direction of Enet is at an angle from radial direction, then = tan 1 E Er 1 2 = tan Thus the inclination of net electric field at point O is ( + ) At a point on the axis of a dipole : q Electric field due to +q charge E1 = PHYSICS WALLAH q O kq (r ) 2 E2 E1 r Electric field due to q charge E2 = kq (r + ) 2 33 ELECTROSTATICS E = E1 E2 = Net electric field 2kpr E= 2 2 ) (r ) 2 kq (r + ) 2 = If r >>> then E = kq 4r (r 2 2 2 ) [ p = q 2 = Dipole moment] 2kp r3 At a point on equitorial line of dipole : kq Electric field due to +q charge E1 = x2 ; Electric field due to q kq charge E 2 = E1 sin E1 x2 P E Vertical component of E1 and E2 will cancel each other and horizontal components will be added So net electric field at P E2 x E= 2kq x 3 = 2kq x 2 cos cos = 2kq (r + 2 = 2 3 2 ) kp (r + 2 x r E = E1cos + E2cos [ E1 = E2] E = 2E1 cos = E2 sin (r 2 kq 2 3 2 ) x and x = r 2 + q 2 If r > > > then E = kp r 3 O q kp or E = r3 GOLDEN KEY POINTS For a dipole, potential is zero at equatorial position, while at any finite point E 0 In a uniform E , dipole may feel a torque but not a force. If a dipole placed in a field E (Non-Uniform) generated by a point charge, then torque on dipole may be zero, but F 0 Distribution Point charge Dipole 1 2 Potential proportional to r r E proportional to r 2 r 3 Force between Point charge Proportional to r 2 Dipole and Dipole-dipole point charge r 3 r 4 Example A short electric dipole is situated at the origin of coordinate axis with its axis along x axis and equator along y axis. It is found that the magnitudes of the electric intensity and electric potential due to the dipole are equal at a point distant r = vector of the point in first quadrant. Solution E P = VP cos = 1 2 kp r 3 1 + 3cos 2 = kp cos 1+3 cos2 = 5 cos2 = 45 Position vector r of point P is r = PHYSICS WALLAH r2 5 m from origin. Find the position 5 (i + j) 2 34 ELECTROSTATICS Example Prove that the frequency of oscillation of an electric dipole of moment p and rotational inertia I for small amplitudes about its equilibrium position in a uniform electric field strength E is pE 2 I 1 Solution Let an electric dipole (charge q and q at a distance 2a apart) placed in a uniform external electric field of strength E. Restoring torque on dipole = pE sin = pE (as is small) Here ve sign shows the restoring tendency of torque. = I angular acceleration = = For SHM a = 2 = I PE I 2 F 2a E P pE comparing we get = Thus frequency of oscillations of dipole n = +q = F -q I 1 2 pE I Do yourself - 5 : (i) Three charges are placed at the vertices of an equilateral triangle as shown. Find the dipole moment of the system. y +2q a q (ii) a (0,0) q x Find the Electric field due to the dipole ( p ) at a point 'A' as shown in figure: y P A x (r, 0) (iii) A dipole ( p ) is placed in a uniform electric field aligned parallel to it. Find work required to be done in rotating the dipole by 1800. E P PHYSICS WALLAH 35 ELECTROSTATICS ELECTROSTATIC PRESSURE Force due to electrostatic pressure is directed normally outwards to the surface . Force on small element ds of charged conductor dF = (Charge on ds ) x Electric field = ( ds) Inside + ++ ds ++ + + + + + ++ + + + + 2 0 = 2 2 0 ds E1 E2 = 0 E1 = E2 E1 E2 E2 E1 E2 = Just outside E = E1 + E2 = 2E2 + 2 0 (E1 is field due to point charge on the surface and E2 is field due to rest of the sphere). The electric force acting per unit area of charged surface is defined as electrostatic pressure. Peleectrostatic = dF dS = 2 2 0 Equilibrium of liquid charged surfaces (Soap bubble) Pressures (forces) act on a charged soap bubble , due to (i) Surface tension PT (inward) (ii) Air outside the bubble Po (inward) (iii) Electrostatic pressure Pe (outward) (iv) Air inside the bubble Pi (outward) in state of equilibrium inward pressure = + + + ++ + + + + outward pressure PT + Po = Pi +Pe R+ + + ++ Excess pressure of air inside the bubble (Pex) = Pi Po = PT Pe but PT = 4T r and Pe = 2 2 0 Pex = 4T r 2 if Pi = Po then 2 0 4T r = ++ + + 2 2 0 Example Brass has a tensile strength 3.5 108 N/m2. What charge density on this material will be enough to break it by electrostatic force of repulsion? How many excess electrons per square will there then be? What is the value of intensity just out side the surface? Solution We know that electrostatic force on a charged conductor is given by So the conductor will break by this force if, 2 2 0 dF ds = 2 2 0 > Breaking strength i.e., 2 > 2 9 10 12 3.5 108 i.e. min = ( 3 7 ) 10 2 = 7.94 10 2 ( C / m 2 ) Now as the charge on an electron is 1.6 10 19 C, the excess electrons per m2 Further as in case of a conductor near its surface E = PHYSICS WALLAH 0 = 7.94 10 2 9 10 12 = 8.8 10 9 V/m 36 ELECTROSTATICS CONDUCTOR AND IT'S PROPERTIES [FOR ELECTROSTATIC CONDITION] (i) Conductors are materials which contains large number of free electrons which can move freely inside the conductor. (ii) n electrostatics, conductors are always equipotential surfaces. (iii) Charge always resides on outer surface of conductor. (iv) f there is a cavity inside the conductor having no charge then charge will always reside only on outer surface of conductor. (v) Electric field is always perpendicular to conducting surface. (vi) Electric lines of force never enter into conductors. (vii) Electric field intensity near the conducting surface is given by formula = E = EA = A 0 n ; E B = B 0 n and EC = C 0 0 n n (viii) When a conductor is grounded its potential becomes zero. (ix) (x) When an isolated conductor is grounded then its charge becomes zero. When two conductors are connected there will be charge flow till their potential becomes equal. (xi) Electric pressure at the surface of a conductor is given by formula P = 2 2 0 where is the local surface charge density. Example Prove that if an isolated (isolated means no charges are near the sheet) large conducting sheet s given a charge then the charge distributes equally on its two surfaces. Solution Let there is x charge on left side of sheet and Q x charge on right side of sheet. x Q-x Since point P lies inside the conductor so EP = O x 2 A O Q x 2A O =0 2x 2A O = Q 2A O x= Q 2 Q x= P Q-x 2A 0 Q x 2A 0 2 So charge is equally distributed on both sides Example If an isolated infinite sheet contains charge Q1 on its one surface and charge Q2 on its other surface then prove that electric field intensity at a point in front of sheet will be Q 2A O , where Q = Q1 + Q2 Solution Electric field at point P : E = E Q1 + E Q 2 = Q1 2A 0 n + Q2 2A 0 n = Q1 + Q 2 2A 0 n = Q 2A 0 n [This shows that the resultant field due to a sheet depends only on the total charge of the sheet and not on the distribution of charge on individual surfaces]. PHYSICS WALLAH 37 ELECTROSTATICS Example Three large conducting sheets placed parallel to each other at finite distance contains charges Q, 2Q and 3Q respectively. Find electric field at points A, B , C, and D Q A Sol. -2Q B 3Q C EA = EQ + E 2Q + E3Q. (i) Here EQ means electric field due to Q . EA = (ii) EB 2A 0 = (iii) EC = (iv) (Q 2Q + 3Q) = 2A 0 Q ( 2Q + 3Q) 2A 0 (Q 2Q) (3Q) ED = 2Q 2A 0 = (Q 2Q + 3Q) 2A 0 = Q A 0 , towards left , towards right = 0 4Q = 2A 0 = 2Q 2A 0 2Q A 0 = , towards right Q A 0 2Q A 0 towards left , towards right Example Two conducting plates A and B are placed parallel to each other. A is given a charge Q1 and B a charge Q2. Prove that the charges on the inner facing surfaces are of equal magnitude and opposite sign. Also find the charges on inner & outer surfaces. Solution Consider a Gaussian surface as shown in figure. Two faces of this closed surface lie completely inside the conductor where the electric field is zero. The flux through these faces is, therefore, zero. The other parts of the closed surface which are outside the conductor are parallel to the electric field and hence the flux on these parts is also zero. The total flux of the electric field through the closed surface is, therefore zero. From Gauss s law, the total charge inside this closed surface should be zero. The charge on the inner surface of A should be equal and opposite to that on the inner surface of B. Q1 q E=0 Q1 A E +q A q B B Q2 P E=0 Q2+q The distribution should be like the one shown in figure. To find the value of q, consider the field at a point P inside the plate A. Suppose, the surface area of the plate (one side) is A. Using the equation E = / (2 0), the electric field at P PHYSICS WALLAH 38 ELECTROSTATICS due to the charge Q1 q = due to the charge q = Q1 q 2A 0 q 2A 0 (downward); due to the charge + q = q 2A 0 (downward), and due to the charge Q2 + q = (upward), Q2 + q 2A 0 (upward). The net electric field at P due to all the four charged surfaces is (in the downward direction) Ep = Q1 q 2A 0 q 2A 0 q + 2A 0 Q2 + q 2A 0 As the point P is inside the conductor, this field should be zero. Hence, Q1 q q + q Q2 q = 0 q = Q1 Q 2 2 This result is a special case of the following result. When charged conducting plates are placed parallel to each other, the two outermost, surfaces get equal charges and the facing surfaces get equal and opposite charges. Example Figure shows three large metallic plates with charges Q, 3Q and Q respectively. Determine the final charges on all the surfaces. Q 3Q Q Solution We assume that charge on surface 2 is x. Following conservation of charge, we see that surfaces 1 has charge ( Q x). The electric field inside the metal plate is zero so fields at P is zero. 1 2 3 3Q 4 5 Q 6 x -Q-x P Resultant field at P : EP = 0 Q x 2A 0 = x + 3Q + Q 2A 0 Q x = x + 4Q x = 5Q 2 Note : We see that charges on the facing surfaces of the plates are of equal magnitude and opposite sign. This can be in general proved by gauss theorem also. Remember this it is important result. Thus the final charge distribution on all the surfaces is : + PHYSICS WALLAH 3Q 2 - 5Q 2 5Q 2 +Q 2 -Q 2 +3Q 2 39 ELECTROSTATICS Example An isolated conducting sheet of area A and carrying a charge Q is placed in a uniform electric field E, such that electric field is perpendicular to sheet and covers all the sheet. Find out charges appearing on its two surfaces. Also Q E Solution Let there is x charge on left side of plate and Q x charge on right side of plate EP = 0 x 2A 0 +E= Q x 2A 0 So charge on one side is Q 2 x A 0 = Q 2A 0 E x = EA o and other side Q 2 Q 2 EA 0 The resultant electric field on the left and right side of the plate. On right side E = Q 2 A 0 Q 2 A 0 Q x x + EA o Q-x 2A 0 P x 2A 0 +E + E towards right and on left side E towards left. SOME OTHER IMPORTANT RESULTS FOR A CLOSED CONDUCTOR. (i) f a charge q is kept in the cavity then q will be induced on the inner surface and +q will be induced on the outer surface of the conductor (it can be proved using gauss theorem) (ii) If a charge q is kept inside the cavity of a conductor and conductor is given a charge Q then q charge will be induced on inner surface and total charge on the outer surface will be q + Q. (it can be proved using gauss theorem) +q+Q q q (iii) Resultant field, due to q (which is inside the cavity) and induced charge on S 1, at any point outside S1 (like B, C) is zero. Resultant field due to q + Q on S2 and any other charge outside S2 , at any point inside of surface S2 (like A, B) is zero S2 .B .C S1 .q q PHYSICS WALLAH .A q+Q 40 ELECTROSTATICS (iv) Resultant field in a charge free cavity in a closed conductor is zero. There can be charges outside the conductor and on the surface also. Then also this result is true. No charge will be induced on the inner most surface of the conductor. No charge (v) Charge distribution for different types of cavities in conductors S2 S2 S1 C (A) q S1 C q (B) charge is at the common centre (S1, S2 spherical) charge is not at the common centre (S1, S2 spherical) S2 S2 (C) q S1 C S1 (D) charge is at the centre of S2 (S2 spherical) charge is not at the centre of S2 (S2 spherical) S2 S2 S1 q C (E) charge is at the centre of S1 (Spherical) C q C S1 q (F) charge not at the centre of S1(Spherical) Using the result that E res in the conducting material should be zero and using result (iii) We can show that Note : In all cases charge on inner surface S1 = q and on outer surface S2 = q. The distribution of charge on S1 will not change even if some charges are kept outside the conductor (i.e. outside the surface S2). But the charge distribution on S2 may change if some charges(s) is/are kept outside the conductor. PHYSICS WALLAH 41 ELECTROSTATICS Example An uncharged conductor of inner radius R1 and outer radius R2 contains a point charge q at the centre as shown in figure S2 S1 q R1 O A C R2 B (i) Find E and V at points A, B and C (ii) If a point charge Q is kept out side the sphere at a distance r (>> R 2) from centre then find out resultant force on charge Q and charge q. +q Solution At point A : VA = At point B : VB = (ii) Kq OA Kq OB + + Kq R2 + K( q) K( q) OB R1 + Kq R2 Force on point charge Q : FQ = , EA = = Kq R2 KqQ r2 Kq OA 3 q q OA , EB = 0; At point C : VC = Kq OC , EC = Kq OC 3 OC r (r = distance of Q from centre O ) Force on point charge q: Fq = 0 (using result (iii) & charge on S1 uniform) Example An uncharged conductor of inner radius R1 and outer radius R2 contains a point charge q placed at point P (not at the centre) as shown in figure? Find out the following : B A R1 P q C S1 R2 D S2 (i) VC (ii) VA (iii) VB (iv) EA (v) EB (vi) force on charge Q if it is placed at B Solution (i) VC = (iii) VB = Kq CP + K( q) R1 + Kq R2 Kq (ii) VA = R2 Kq CB (iv) EA = O (point is inside metallic conductor) (v) EB = Kq CB 2 PHYSICS WALLAH ^ CB (vi) FQ = KQq CB 2 ^ CB 42 ELECTROSTATICS (vi) Sharing of charges : Two conducting hollow spherical shells of radii R1 and R2 having charges Q1 and Q2 respectively and separated by large distance, are joined by a conducting wire. Let final charges on spheres are q1 and q2 respectively. q1 q2 R2 R1 Potential on both spherical shell become equal after joining, therefore Kq1 R1 = Kq 2 R2 ; q1 q2 = R1 ...(i) and R2 q1 + q2 = Q1 + Q2 (Q1 + Q 2 )R 1 ; q2 = from (i) and (ii) q1 = ratio of charges R 1 1 4 R 12 R 1 = ; = q 2 R 2 2 4 R 22 R 2 R1 + R 2 ......(ii) (Q1 + Q 2 )R 2 R1 + R 2 q1 1 ratio of surface charge densities 2 q1 Ratio of final charges q2 1 Ratio of final surface charge densities. = R2 = R1 R1 R2 = 2 R2 R1 Example The two conducting spherical shells are joined by a conducting wire and cut after some time when charge stops flowing. Find out the charge on each sphere after that. Q -3Q R 2R Solution After cutting the wire, the potential of both the shells is equal Thus, potential of inner shell Vin = and potential of outer shell Vout = KR As Vout = Vin 2Q = x 2Q x = 0 R = Kx R Kx 2R + + K( 2Q x) = 2R K( 2Q x) 2R = k ( x 2Q ) 2R KQ R K ( x 2Q ) 2R So charge on inner spherical shell = 0 and outer spherical shell = 2Q. PHYSICS WALLAH 43 ELECTROSTATICS Example Find charge on each spherical shell after joining the inner most shell and outer most shell by a conducting wire. Also find charges on each surface. Q -3Q R 2R 5Q 3 Solution -2Q Q 2 1 R Let the charge on the innermost sphere be x. 2R 3R Finally potential of shell 1 = Potential of shell 3 Kx R + K( 2Q) 2R + K(6Q x) 3R = KQ 3R + k ( 2q ) 3R + k ( 5Q ) 6Q x 3 3R -2Q x 2 1 3x 3Q + 6Q x = 4Q ; 2x = Q ; x= R Q 2 2R 3R Charge on innermost shell = Q 2 , charge on outermost shell = 5Q Q +3Q/ 2 3Q/ 2 Q/ 2 Q/ 2 2 middle shell = 2Q Final charge distribution is as shown in figure. Example Two conducting hollow spherical shells of radii R and 2R carry charges Q and 3Q respectively. How much charge will flow into the earth if inner shell is grounded ? 3Q -Q R 2R Solution When inner shell is grounded to the Earth then the potential of inner shell will become zero 3Q because potential of the Earth is taken to be zero. x Kx R x= = + K3Q 2R 3Q 2 3Q 2 =0 , the charge that has increased ( Q) = PHYSICS WALLAH R Q 2 hence charge flows into the Earth = 2R Q 2 44 ELECTROSTATICS Example An isolated conducting sphere of charge Q and radius R is connected to a similar uncharged sphere (kept at a large distance) by using a high resistance wire. After a long time what is the amount of heat loss ? Solution When two conducting spheres of equal radius are connected charge is equally distributed on them. So we can say that heat loss of system Q2 Q2 / 4 Q2 / 4 Q2 0 + = H = Ui Uf = 8 0 R 8 0 R 8 0 R 16 0 R Do yourself - 6 : (i) Two metallic plates are placed at a small separation d. Left plate is given charge 2Q. Find the final charges appearing on the surfaces 1,2,3 and 4 respectively. 2Q 2 1 (ii) 3 4 A neutral spherical conductor with centre at origin has a spherical cavity with center at coordinates (a, b) as shown in figure. A charge q is placed at the centre of cavity. Find the Electric field at a point A at (r, 0). y r1 q r2 (a,b) (0,0) (iii) A x (r,0) Two conducting hollow spherical shells of radii R and 2R carry charges Q and 4Q respectively. How much charge will flow into the earth if outer shell is grounded? 4Q Q R 2R PHYSICS WALLAH 45 ELECTROSTATICS Answers for Do yourself Do yourself 1 : (i) 3KQq a 2 downwards, (ii) d from Q 3 Do yourself 2 : (i) 4 2KQ i a2 (ii) KQ a 1 ( 2+ ) 2 Do yourself 3 : (i) q (ii) zero 2 0 (iii) 2R 0 Do yourself 4 : (i) Kq 2 a (4 + 2 ) (ii) v = Kq 2 mr (iii) 14xy i 2 j + 3z2 x k Do yourself 5 : (i) qa 3 j (ii) 2kp cos kp sin i j r3 r3 (iii) 2pE Do yourself 6 : (i) Q, Q, Q, Q PHYSICS WALLAH (ii) kq r2 (iii) 3Q 46 ELECTROSTATICS SOME WORKED OUT EXAMPLES Example#1 For a spherically symmetrical charge distribution, electric field at a distance r from the centre of sphere is E = kr 7 r , where k is a constant. What will be the volume charge density at a distance r from the centre of sphere ? (A) = 9k 0r6 (B) = 5k 0r3 (C) = 3k 0r4 (D) =9k 0r0 Solution : Ans. (A) ( E ) ( 4 r 0 0 (Note : Check dimensionally that r6) ( kr ) ( 4 r 7 ( 4 r dr ) )= k r sian Surfac us e E dS = ( 4 r dr ) )= 2 2 Ga By using Gauss law q O dr 2 2 0 9 0 = r 2 dr Example#2 Two positrons (e+) and two protons (p) are kept on four corners of a square of side a as shown in figure. The mass of proton is much larger than the mass of positron. Let q denotes the charge on the proton as well as the positron then the kinetic energies of one of the positrons and one of the protons respectively after a very long time will be (A) (C) e+ p p e+ 1 q2 1 1 + , 1 + 4 0 a 2 2 4 0 a 2 2 q2 q2 q2 , q2 (B) q2 2 0 a 4 2 0 a 1 q2 1 + , 2 0 a 4 2 8 2 0 a q2 (D) 4 0 a 4 0 a , Solution : Ans. (D) As mass of proton >>> mass of positron so initial acceleration of positron is much larger than proton. Therefore positron reach far away in very short time as compare to proton. a e+ a p 4kq 2 2kq 2 2K e + = + a 2 a 2K p = kq 2 a 2 a a e+ p p a a kq 2 q2 1 K = + 1 + and e 2 0 a 4 2 a 2 0 Kp = PHYSICS WALLAH a a p q2 8 2 0 a 47 ELECTROSTATICS Example#3 Four charges are placed at the circumference of a dial clock as shown in figure. If the clock has only hour hand, then the resultant force on a charge q0 placed at the centre, points in the direction which shows the time as : +q 12 q 9 3 q0 +q 6 q (A) 1:30 (B) 7:30 (C) 4:30 (D) 10:30 Solution : Ans. (B) +q q 9 +q q0 7. 30 Fnet 6 q Example#4 A small electric dipole is placed at origin with its dipole moment directed along positive xaxis. The direction of electric field at point (2, 2 2,0) is (A) along z-axis (B) along y-axis (C) along negative y-axis (D) along negative z-axis Solution : Ans. (B) y (2,2 2) q tan = y x = x +q 2; cot = 1 2 Also tan = tan 2 = 1 = cot 2 + = 90 i.e., E is along positive y-axis. Example#5 Uniform electric field of magnitude 100 V/m in space is directed along the line y = 3 + x. Find the potential difference between point A (3, 1) & B (1,3). (A) 100 V (B) 200 2V (C) 200 V (D) zero Solution : Ans. (D) Slope of line AB = PHYSICS WALLAH 3 1 1 3 = 1 which is perpendicular to direction of electric field. 48 ELECTROSTATICS Example#6 The diagram shows a uniformly charged hemisphere of radius R. It has volume charge density . If the electric field at a point 2R distance above its centre is E then what is the electric field at the point which is 2R below its centre? A B (A) R 6 0 +E (B) R 12 0 E (C) R 6 0 +E (D) R 24 0 +E Solution : Ans. (B) + + Apply principle of superposition Electric field due to a uniformly charged sphere = R 12 0 R ; Eresultant = 12 0 E Example#7 A metallic rod of length l rotates at angular velocity about an axis passing through one end and perpendicular to the rod. If mass of electron is m and its charge is e then the magnitude of potential difference between its two ends is m 2 2 (A) ( 2e ) (B) m 2 2 (C) e m 2 (D) None of these e Solution : Ans. (A) When rod rotates the centripetal acceleration of electron comes from electric field E = mr 2 Thus, V = E.dr = mr 2 e 0 dr = m 2 e 2 e 2e eE E Example#8 Consider a finite charged rod. Electric field at Point P (shown) makes an angle with horizontal dotted line then angle is :- 0 32 0 88 (A) 60 Solution : Ans. B Required angle = PHYSICS WALLAH (B) 28 2 1 2 = (C) 44 88 32 2 = 56 2 (D) information insufficient = 28 49 ELECTROSTATICS Example#9 The electric potential in a region is given by the relation V(x) = 4 + 5x 2. If a dipole is placed at position ( 1, 0) with dipole moment p pointing along positive Y-direction, then (A) Net force on the dipole is zero. (B) Net torque on the dipole is zero (C) Net torque on the dipole is not zero and it is in clockwise direction (D) Net torque on the dipole is not zero and it is in anticlockwise direction Solution : Ans. (AC) F p ( 1,0) (0,0) F V(x) = 4 + 5x2 E = 10xi Net force will be zero and torque not zero and rotation will be along clockwise direction Example#10 to 12 A thin homogeneous rod of mass m and length is free to rotate in vertical plane about a horizontal axle pivoted at one end of the rod. A small ball of mass m and charge q is attached to the opposite end of this rod. The whole system is positioned in a constant horizontal electric field of magnitude E = mg 2q . The rod is released from shown position from rest. m m 10. (A) 11. 8g (B) 9 3g (C) 2 9g (D) 8 2g 9 What is the acceleration of the small ball at the instant of releasing the rod? (A) 12. E What is the angular acceleration of the rod at the instant of releasing the rod? 8g (B) 9 9g (C) 8 7g (D) 8 8g 7 What is the speed of ball when rod becomes vertical? (A) 2g (B) 3 3g 4 (C) 3g 2 4g (D) 3 Solution : 10. Ans. (C) /2 /2 mg mg m 2 ( ) +m By taking torque about hinge I = mg + mg when I = 3 2 PHYSICS WALLAH 2 = 9g 8 50 ELECTROSTATICS 11. Ans. (B) 9 9g = g 8 8 Acceleration of ball = = 12. Ans. (C) From work energy theorem 1 4 m 2 3 2 I 2 = mg + mg qE 2 2 1 mg 2 2 3 m 2 2 = mg = = mg 2 2 3 Speed of ball = = 3g 2 3g 2 Example#13 A simple pendulum is suspended in a lift which is going up with an acceleration of 5 m/s2. An electric field of magnitude 5 N/C and directed vertically upward is also present in the lift . The charge of the bob is 1 C and mass is 1 mg. Taking g = 2 and length of the simple pendulum 1m, find the time period of the simple pendulum (in sec). Solution : Ans. 2 T = 2 g eff geff = g qE + 5 = 15 M 1 5 10 6 E 5m/ s2 1 10 6 geff = 10 = 2 T = 2 sec Example#14 The variation of potential with distance x from a fixed point is shown in figure. Find the magnitude of the electric field (in V/m) at x =13m. V(volt) 60 45 30 20 10 0 2 4 6 8 10 12 14 16 x(m) Solution : Ans. 5 E= dV dx = 20 4 = 5 volt/meter Example#15 The energy density u is plotted against the distance r from the centre of a spherical charge distribution on a log-log scale. Find the magnitude of slope of obtianed straight line. Solution : Ans. 2 2 q q2 q2 log u = log = log k 2 log r u= 0E = 0 = 2 2 2 32 2 0 r 2 2 4 0 r 32 0 r 1 2 PHYSICS WALLAH 1 51 ELECTROSTATICS Example#16 The figure shows four situations in which charges as indicated (q > 0) are fixed on an axis. How many situations is there a point to the left of the charges where an electron would be in equilibrium? +q (1) +4q 4q q q 4q (3) (2) +4q q (4) Solution : Ans. 2 For (1) Let the electron be held at a distance x from +q charge. x e For equilibrium q ( e) 4 0 x 2 d +q 4q ( e )( 4 q ) = 4 0 ( x + d ) 2 We can find value of x for which Fnet = 0 which means that electron will be in equilibrium. x d q e +4q (x+d) For (2) : For equilibrium ( e )( q ) 4 0 x 2 = ( e )4 q 4 0 ( x + d ) 2 We can find value of x for which Fnet = 0 which means that electron will be in equilibrium. In case (3) and (4) the electron will not remain at rest, since it experiences a net non zero force. OR Equilibrium is always found near the smaller charge Example#17 An electric field is given by E = ( yi + xj ) N C . Find the work done (in J) in moving a 1C charge from rA = ( 2i + 2 j) m to rB = ( 4i + j) m . Solution : Ans. 0 A= (2, 2) and B = (4, 1); B B A A WA B = q ( VB VA ) = q dV = qE.dr B B ( 4,1) A ( 2,2 ) = q ( yi + xj ). ( dxi + dyj ) = q ( ydx + xdy ) = q A PHYSICS WALLAH ( d ( xy ) = q xy ( 4,1) 2,2 ) = q 4 4 = 0 52 ELECTROSTATICS Example#18 The arrangement shown consists of three elements. 3 1 2 (i) A thin rod of charge 3.0 C that forms a full circle of radius 6.0 cm. (ii) A second thin rod of charge 2.0 C that forms a circular arc of radius 4.0 cm and concentric with the full circle, subtending an angle of 90 at the centre of the full circle. (iii) An electric dipole with a dipole moment that is perpendicular to a radial line and has magnitude 1.28 10 21C-m. Find the net electric potential in volts at the centre. Solution Ans. (0) Potential due to dipole at the centre of the circle is zero. Potentials due to charge on circle = V1 = K.( 3 10 6 ) 6 10 2 K.(2 10 6 ) Potential due to arc V2 = 4 10 2 Net potential = V1 + V2= 0 Example#19 Six charges are kept at the vertices of a regular hexagon as shown in the figure. If magnitude of force applied by +Q on +q charge is F, then net electric force on the +Q is nF. Find the value of n. +q +q +3q -2q +Q -3q -4q Solution : Ans. 9 +q +q +3q 5F 5F 4F -2q -3q -4q Fnet = 9F PHYSICS WALLAH 53 ELECTROSTATICS Example#20 Electric field in a region is given by E = 4 xi + 6 yj . The charge enclosed in the cube of side 1m oriented as shown in the diagram is given by 0. Find the value of . z y x Solution Ans. 2 z y x = (6y) Area (4x) Area = 6 1 (1)2 4 1 (1)2 = 2 therefore q 0 = 2 q = 2 0 Example#21 An infinite plane of charge with = 2 0 C m2 is tilted at a 37 angle to the vertical direction as shown below. Find the potential difference, VA VB in volts, between points A and B at 5 m distance apart. (where B is vertically above A). B A Solution Ans. 3 B 53 37 A PHYSICS WALLAH E= 2 0 = 1 N/C VB VA = E.d = 2 0 ( ) 5 cos 53 = 3V 54 ELECTROSTATICS EXERCISE #1 HCV Worked out Examples (Chapter No. 29 - 1,2,3,4,5,6) 1. Two particles of charges 2q and q are fixed at points A and B, distance apart. Where should a positive test charge be placed on the line connecting the charges for it to be in equilibrium? What is the nature of the equilibrium for small disturbances along the line joining the charges. 2. Six charges are kept at the vertices of a regular hexagon as shown in the figure. If magnitude of force applied by +Q on +q charge is F, then net electric force on the +Q is nF. Find the value of n. +q 2q +q +Q 3q +3q 4q 3. Three charges 4q, Q and q are in a straight line in the position of 0, l/2 and l respectively. The resultant force on q will be zero, if Q = HCV Exercises (Chapter No. 29 - 3,5,6,7,12,15,16,18,19,20,26,27,31,32,33) HCV Worked out Examples (Chapter No. 29 - 8,9,11,12,13) 4. Draw E-x graph for 0 < x < x0, for a system of two point charges A and B. Charge A is kept at the origin and B at (x0, 0), when (i) both are positive. (ii) both are negative. (iii) A is positive and B is negative. (iv) A is negative and B is positive. 5. A clock face has negative charges q, 2q, 3q, ........., 12q fixed at the positions of the corresponding numerals on the dial. Assume that the clock hands do not disturb the net field due to point charges. At what time does the hour hand point in the same direction is electric field at the centre of the dial. 6. A charge +10 9 C is located at the origin in free space and another charge Q at (2, 0, 0). If the X-component of the electric field at (3, 1, 1) is zero, calculate the value of Q. Is the Y-component zero at (3, 1, 1)? 7. Two mutually perpendicular infinite wires along x-axis and y-axis carry charge densities 1 and 2. The electric line of force at P is along the line y = 1 x , where P is also a point lying 3 on the same line then find 1/ 2 2 1 PHYSICS WALLAH 55 ELECTROSTATICS 8. A particle of mass m and negative charge q is thrown in a gravity free space with speed u from the point A on the large charged sheet of uniform surface charge density , as shown in figure. Find the maximum distance from A on sheet where the particle can strike. + + + + + + + + A + 9. u Calculate the magnitude of electrostatic force on a charge placed at a vertex of a triangular pyramid (4 vertices, 4 faces), if 4 equal charges are placed at all four vertices of pyramid of side a . HCV Exercises (Chapter No. 29 - 34,36,37,39,40,41,42,43,44,50,51,52) HCV Worked out Examples(Chapter No. 30 - 1,2,3,6) 10. The length of each side of a cubical closed surface is . If charge q is situated on one of the vertices of the cube, then find the flux passing through shaded face of the cube. q 11. A point charge Q is located on the axis of a disc of radius R at a distance a from the plane of the disc. If one fourth (1/4th) of the total flux from the charge passes through the disc, find the relation between a & R. R a Q 12. Consider a triangular surface whose vertices are three points having co ordinate A(2a, 0, 0), B(0, a, 0), C(0, 0, a). If there is a uniform E field E i + 2 E j + 3E k then flux linked to 0 triangular surface ABC is 13. E0 a 2 0 0 2 N . Find the value of N. What is the flux (in SI units) through the part of surface x 2 + y2 + z2 =1 for region x > 0, y > 0, 8 N / C j . z > 0, due to the electric field E = 14. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8 103 Nm2/C.(here 0 = 8.85 10 12 C2/N-m2) (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not? PHYSICS WALLAH 56 ELECTROSTATICS 15. An electric field is given by E = 4i + 3 ( y 2 + 2 ) j pierces gaussian cube of side 1m placed at origin such that one of its corners is at origin & rest of sides are along positive side of coordinate axis. If the magnitude of net charge enclosed is n 0 then n (in SI units) will be equal to. HCV Exercises (Chapter No. 30 - 1,2,3,4,5,6,7,8) HCV Worked out Examples(Chapter No. 29 - 14,15,16,17) 16. Consider the configuration of a system of four charges each of value +q. Find the work done by external agent in changing the configuration of the system from figure (i) to figure (ii). +q +q +q a a +q +q +q +q a +q fig (i) fig (ii) 17. Three charges 0.1 coulomb each are placed on the corners of an equilateral triangle of side 1 m. If the energy is supplied to this system at the rate of 1 kW, how much time would be required to move one of the charges onto the midpoint of the line joining the other two? 18. A point charge +q and mass 100 gm experiences a force of 100 N at a distance 20 cm from a straight infinitely long uniformly charged straight wire. If the charge particle is released there, find its speed when it is at a distance 40 cm from wire. 19. A charge + Q is uniformly distributed over a thin ring of radius R and very large mass. A particle of charge Q and mass m starts from rest at a point on the axis of the ring far away from the centre of the ring. Find the velocity of this particle at the moment it passes through the centre of the ring. 20. A small ball of mass 2 10 3 Kg having a charge of 1 C is suspended by a string of length 0. 8 m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball so that it can make complete revolution. 21. A simple pendulum of length and bob mass m is hanging in front of a large nonconducting sheet of surface charge density . If suddenly a charge +q is given to the bob in the position shown in figure. Find the maximum angle through which the string is deflected from vertical. + + + + + + + + + PHYSICS WALLAH 57 ELECTROSTATICS 22. There are 27 drops of a conducting fluid. Each has radius r and they are charged to a potential V0. They are then combined to form a bigger drop. Find its potential. 23. The electric field strength depends only on the x, y and z coordinates according to the law E= a(xi + yj + zk) (x 2 + y 2 + z 2 ) 3/ 2 , where a = 122.5 SI unit and is a constant. Find the potential difference (in volt) between (3, 2, 6) and (0, 3, 4). HCV Exercises (Chapter No. 29 - 35,38,47,48,53,56,57,58,60,61,62,63,64) HCV Worked out Examples (Chapter No. 29 - 18,19, Chapter No. 30 - 5) 24. An electric dipole placed along x-axis with its dipole moment vector pointing along positive xaxis, is placed at the centre of wire frame as shown in figure. A fixed circular wire frame lies in y-z plane with 20 identical charges of magnitude q, kept symmetrically fixed to wire frame. What is work done (in joule) by an external agent in slowly turning dipole through 180 from initial orientation of dipole moment vector. y +q +q Dipole q z x +q +q Non conducting ring +q Radius of ring : R = 1 cm Length of dipole : r = 1 mm Charged : q = 1 C 25. A dipole is placed at origin of coordinate system as shown in figure, find the electric field at point P(0, y). y P (0, y) p 26. 45 x A dipole of dipole moment p = 2i 3j + 4k is placed at point A(2, 3,1). The electric potential due to this dipole at the point B(4, 1,0) is (ab) 109 volt here 'a' represents sign (for negative answer select 0 for positive answer select 1. Write the value of (a+b)2. The parameters specified here are in S.I. units. HCV Exercises (Chapter No. 29 -70,71,72,73,74) HCV Worked out Examples (Chapter No. 30 - 4,8) PHYSICS WALLAH 58 ELECTROSTATICS 27. A particle of mass m and charge q moves along a diameter of a uniformly charged sphere of radius R and carrying a total charge + Q. Find the frequency of SHM of the particle if the amplitude does not exceed R. 28. Two thin conducting shells of radii R and 3R are shown in figure. The outer shell carries a charge +Q and the inner shell is neutral. The inner shell is earthed with the help of switch S. Find the charge attained by the inner shell. 3R R 29. +Q S Consider two concentric conducting spheres of radii a & b (b > a). Inside sphere has a positive charge q1. What charge should be given to the outer sphere so that potential of the inner sphere becomes zero? How does the potential vary between the two spheres & outside ? 30. Consider three identical metal spheres A, B and C. Spheres A carries charge + 6q and sphere B carries charge 3q. Sphere C carries no charge. Spheres A and B are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Finally the sphere C is touched to sphere B and separated from it. Find the final charge on the sphere C. HCV Exercises (Chapter No. 29 - 45, Chapter No. 30 - 9,10,11,12,13,16,17,21,22,23,24) PHYSICS WALLAH 59 ELECTROSTATICS EXERCISE #2 1. Six charges are placed at the vertices of a regular hexagon as shown in the figure. Find the electric field on the line passing through O and perpendicular to the plane of the figure as a function of distance x from point O. Q +Q Q +Q O Q +Q a 2. A circular ring of radius R with uniform positive charge density per unit length is fixed in the Y-Z plane with its centre at the origin O. A particle of mass m and positive charge q is projected from the point P ( 3 R, 0, 0) on the positive X-axis directly towards O, with initial velocity v. Find the smallest value of the speed v such that the particle does not return to P. 3. Two small balls having the same mass & charge & located on the same vertical at heights h 1 & h2 are thrown in the same direction along the horizontal at the same velocity v. The 1st ball touches the ground at a distance from the initial vertical . At what height will the 2nd ball be at this instant ? The air drag & the charges induced should be neglected. 4. Two concentric rings of radii r and 2r are placed with centre at origin. Two charges +q each are fixed at the diametrically opposite points of the rings as shown in figure. Smaller ring is now rotated by an angle 90 about Z-axis then it is again rotated by 90 about Y-axis. Find the work done by electrostatic forces in each step. If finally larger ring is rotated by 90 about X-axis, find the total work required to perform all three steps. y +q +q x +q z 5. +q Small identical balls with equal charges are fixed at vertices of regular 2008 - gon with side a. At a certain instant, one of the balls is released & a sufficiently long time interval later, the ball adjacent to the first released ball is freed. The kinetic energies of the released balls are found to differ by K at a sufficiently long distance from the polygon. Determine the charge q of each ball. PHYSICS WALLAH 60 ELECTROSTATICS 6. A nonconducting ring of mass m and radius R is charged as shown. The charged density i.e. charged per unit length is. It is then placed on a rough nonconducting horizontal surface plane. At time t = 0, a uniform electric field E = E 0 i is switched on and the ring start rolling without sliding. Determine the friction force (magnitude and direction) acting on the ring, when it starts moving. y x 7. Find the electric field at centre of semicircular ring shown in figure. Y q q + R 8. + x A cone made of insulating material has a total charge Q spread uniformly over its sloping surface. Calculate the energy required to take a test charge q from infinity to apex A of cone. The slant length is L. A AB = L B 9. Two concentric rings, one of radius a and the other of radius b have the charges +q and (2/5) 3/2q respectively as shown in the figure. Find the ratio b/a if a charge particle placed on the axis at z = a is in equilibrium. 3/2 qB = (2/5) q b a z =a qA=+q 10. A positive charge Q is uniformly distributed throughout the volume of a nonconducting sphere of radius R. A point mass having charge +q and mass m is fired towards the centre of the sphere with velocity v from a point at distance r(r > R) from the centre of the sphere. Find the minimum velocity v so that it can penetrate R/2 distance of the sphere. Neglect any resistance other than electric interaction. Charge on the small mass remains constant throughout the motion. PHYSICS WALLAH 61 ELECTROSTATICS 11. A cavity of radius r is present inside a solid dielectric sphere of radius R, having a volume charge density of . The distance between the centres of the sphere and the cavity is a. An electron e is released inside the cavity at an angle = 45 as shown. How long will it take to touch the sphere again? r R 12. a e A non-conducting disc of radius a and uniform positive surface charge density is placed on the ground, with its axis vertical. A particle of mass m & positive charge q is dropped, along the axis of the disc, from a height H with zero initial velocity. The particle has q m = 4 0 g . (a) Find the value of H if the particle just reaches the disc. (b) Sketch the potential energy of the particle as a function of its height and find itsc equilibrium position. 13. A nonuniform but spherically symmetric distribution of charge has a charge density given as follow: = 0 (1 r/R) for r R, =0 for r R, where 0 = 3Q / R3 is a constant. (a) Show that the total charge contained in the charge distribution is Q. (b) Show that, for the region defined by r R, the electric field is identical to that produced by a point charge Q. (c) Obtain an expression for the electric field in the region r R. (d) Compare your results in parts (b) and (c) for r = R. 14. Four point charges +8 C, 1 C, 1 C and + 8 C are fixed at the points + 3 2 m and + 27 2 27 2 m, 3 2 m, m respectively on the y axis. A particle of mass 6 10 4 kg and charge +0.1 C moves along the x direction. Its speed at x = + is v0. Find the least value of v0 for which the particle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free. Given : 1/(4 0 ) = 9 109 Nm2/C2). PHYSICS WALLAH 62 ELECTROSTATICS 15. (a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (E 2 E 1 ).n = 0 where n is a unit vector normal to the surface at a point and is the surface charge density at that point. (The direction of n is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is n 0 (b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint : For (a), use Gauss s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero. 16. A spherical shell has uniform charge density 8.8 10 11 C/m2. If a pin hole is made in the surface of the shell then find the electric field in the hole in N/C. (Take 0 = 8.8 10 12 S.I. units) 17. Can the potential function have a maximum or minimum in free space ? 18. Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume. 19. If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero. 20. A thick shell with inner radius R and outer radius 3R has uniform charge density c/m3. It has a spherical cavity of radius R as shown in the figure. The electric field at the centre O 1 of the cavity is PHYSICS WALLAH +++ +++++ ++++++ ++++++ +++++++ +++++++ ++++++ ++++++ +++++ +++ +++ +++++ ++++++ ++++++ +++++++ + + + + + +O+ ++++++ ++++++ +++++ +++ O1 63 ELECTROSTATICS EXERCISE #3 1. Mid way between the two equal and similar charges, we placed the third equal and similar charge. Which of the following statements is correct, concerned to the equilibrium along the line joining the charges ? (A) The third charge experienced a net force inclined to the line joining the charges (B) The third charge is in stable equilibrium (C) The third charge is in unstable equilibrium (D) The third charge experiences a net force perpendicular to the line joining the charges 2. Four charges are arranged at the corners of a square ABCD, as shown. The force on a +ve charge kept at the centre of the square is B A +q q 2q +2q (A) zero (C) along diagonal BD 3. C D (B) along the diagonal AC (D) perpendicular to the side AB Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at its centre O, force on it will be: A B q 3q 2q F 2q C O q 2q D E (A) Zero (B) Along OF (C) Along OC (D) None of these 4. Two equal negative charges are fixed at the points [0, a ] and [0, a] on the y-axis. A positive charge Q is released from rest at the points [2a, 0] on the x-axis . The charge Q will (A) move to the origin and stop there (B) move to the origin, cross the origin and then move to infinitely. (C) execute simple harmonic motion about the origin. (D) execute oscillatory but not simple harmonic motion. 5. Two free positive charges 4q and q are a distance apart. What charge Q is needed to achieve equilibrium for the entire system and where should it be placed form charge q? (A) Q = (B) Q = (C) Q = (D) Q = 4q 9 4q 9 4q 9 4q 9 in between 4q and q at in between 4q and q at in between 4q and q at in between 4q and q at PHYSICS WALLAH 3 2 3 3 2 3 from 4q from 4q from 4q from 4q 64 ELECTROSTATICS 6. Two identical positive charges are fixed on the y-axis, at equal distances from the origin O. A particle with a negative charge starts on the x-axis at a large distance from O, moves along the + x-axis, passes through O and moves far away from O. Its acceleration a is taken as positive in the positive x-direction. The particle s acceleration a is plotted against its x-coordinate. Which of the following best represents the plot? a x (A) 7. a O a x (B) (C) O a x O (D) O x A small particle of mass m and charge q is placed at point P on the axis of uniformly charged ring and released. If R >> x, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to Q R P x (A) 8. qQ 4 0 mR 3 (B) qQx 4 0 mR (C) 4 qQ 4 0 mR 3 (D) qQx 4 0 mR 4 A wheel having mass m has charges +q and q on diametrically opposite points. It remains in equilibrium on a rough inclined plane in the presence of uniform vertical electric field E = +q E q (A) 9. mg q (B) mg 2q (C) mg tan 2q (D) None of these A point charge 50 C is located in the XY plane at the point of position vector r0 = 2i + 3 j . What is the electric field at the point of position vector r = 8i 5 j (A) 1200 V/m 10. (B) 0.04 V/m (C) 900 V/m (D) 4500 V/m A point charge q is placed at origin. Let E A , E B and E C be the electric field at three points A (1, 2, 3), B (1, 1, 1) and C(2, 2, 2) due to charge q. Consider the following statements and choose correct alternative. (i) E A E B (ii) | E B |= 4 | E C | (A) only [i] is correct (B) only [ii] is correct (C) both [i] and [ii] are correct (D) both [i] and [ii] are wrong PHYSICS WALLAH 65 ELECTROSTATICS 11. Two identical point charges are placed at a separation of L. P is a point on the line joining the charges, at a distance x from any one charge. The field at P is E. E is plotted against x for values of x from close to zero to slightly less than L. Which of the following best represents the resulting curve? E E (C) O L x (D) O E E (A) (B) O L x O L x L x 12. A nonconducting ring of radius R has uniformly distributed positive charge Q. A small part of the ring, of length d, is removed (d << R). The electric field at the centre of the ring will now be (A) directed towards the gap, inversely proportional to R3. (B) directed towards the gap, inversely proportional to R2. (C) directed away from the gap, inversely proportional to R3. (D) directed away from the gap, inversely proportional to R2. 13. In the given figure the direction ( ) of E at point P due to uniformly charged finite thin rod will be y Ey d P + + + + + + 30 + + (A) at an angle of 30 from x-axis (C) at an angle of 60 from x -axis 14. y x Ex (B) at an angle of 45 from x -axis (D) None of these An equilateral triangle wire frame of side L having 3 point charges at its vertices is kept in x-y plane as shown. Centroid of the triangle conicides with the origin. Component of electric field due to the configuration in z direction at (0, 0, L) is y q x q (A) 15. 9 3 kq 2 8L (B) 2q 9kq (C) zero 8L2 (D) None of these A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable (A) for all values of H (B) only if H > R 2 PHYSICS WALLAH (C) only if H < R 2 (D) only if H = R 2 66 ELECTROSTATICS 16. Select the correct statement : (Only force on a particle is due to electric field) (A) A charged particle always moves along the electric line of force. (B) A charged particle may move along the line of force (C) A charge particle never moves along the line of force (D) A charged particle moves along the line of force only if released from rest. 17. The figure shows the electric field lines in the vicinity of two point charges. Which one of the following statements concerning this situation is true? q1 q2 (A) q1 is negative and q2 is positive (B) The magnitude of the ratio (q2/q1) is less than one (C) Both q1 and q2 have the same sign of charge (D) The electric field is strongest midway between the charges. Comprehension for Q. no. 18 and 19 In the figure, electric field lines are shown for an isolated system of two charges A and B. Study the lines of forces shown and answer the following questions. B A 18. From the figure you can certainly conclude that (A) Both of them have same magnitude and opposite sign. (B) Both of them have opposite sign and magnitude of A is greater than that of B. (C) Both of them have opposite sign and magnitude of B is greater than that of A. (D) Both of them have opposite sign but information is insufficient to predict their relative magnitudes. 19. Charges A and B have (A) magnitudes in the ratio 5: 3 respectively (B) magnitudes in the ratio 3:2 respectively (C) magnitudes in the ratio 3:5 respectively (D) magnitudes in the ratio 2:3 respectively 20. Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in (A) PHYSICS WALLAH (B) (C) (D) 67 ELECTROSTATICS 21. The dimensions of (A) MLT 1 22. 1 2 0 E 2 ( 0 : permittivity of free space ; E : electric field) is: (B) ML2T 2 (C) MLT 2 (D) ML 1T 2 Electric flux through a surface of area 100 m2 lying in the xy plane is (in V-m) if E = i + 2 j + 3k (A) 100 23. (B) 141.4 (C) 173.2 (D) 200 A large sheet of uniform charge passes through a hypothetical spherical surface. The figure shows principal section of the situation. The electric flux through the spherical surface is given by R x (A) 24. R 2 0 (B) 2 (R 2 x 2 ) 0 (C) (R x) 2 0 (D) (R 2 x 2 ) 0 A uniform electric field E = ai + bj , intersects a surface of area A. What is the flux through this area if the surface lies in the yz plane? (A) a A 25. (B) 0 (C) b A (D) A a 2 + b 2 The electric field in a region is given by E = 200i N/C for x > 0 and 200i N/C for x < 0. A closed cylinder of length 2m and cross-section area 102 m2 is kept in such a way that the axis of cylinder is along X-axis and its centre coincides with origin. The total charge inside the cylinder is [Take : 0 = 8.85 10 12 C2m2.N) (A) zero 26. (B) 1.86 10 5C (C) 1.77 10 11C (D) 35.4 10 8 C In the given figure flux through surface S1 is 1 & through S2 is 2. Which is correct ? (A) 1 = 2 PHYSICS WALLAH (B) 1 > 2 (C) 1 < 2 (D) None of these 68 ELECTROSTATICS 27. A charge +Q is located somewhere inside a vertical cone such that the depth of the charge from the free surface of the cone is H. It is found that the flux associated with the cone with the curved surface is 3Q 5 0 . If the charge is raised vertically through a height 2H, then the flux through the curved surface is H +Q (A) 28. 3Q 5 0 (B) 2Q (C) 5 0 4Q 5 0 (D) Zero Refer to the arrangement of charges in Fig. and a Gaussian surface of radius R with Q at the centre. Then Gaussian surface Q R 5Q R/2 R 2Q Fig. (A) total flux through the surface of the sphere is (B) field on the surface of the sphere is Q 0 Q 4 0 R 2 (C) flux through the surface of sphere due to Q is zero. (D) field on the surface of sphere due to 2Q is same everywhere. 29. Three uniformly charged wires with linear charge density are placed along x, y and z axis respectively. What is flux of electric field through Gaussian surface given by x 2 + y2 + z2 = 1 ; x>0;y>0;z>0 (A) 30. 3 2 0 (B) 3 0 (C) 6 0 (D) 3 4 0 E varies along x as E = 3x2 the volume charge density at x = 1 is :(A) 6 0 (B) 6 0 (C) 3 0 (D) None of these Electric Potential Energy and Electric Potential 31. Which of the following is a volt : (A) Erg per cm (C) Newton per coulomb PHYSICS WALLAH (B) Joule per coulomb (D) Newton per coulomb per m2 69 ELECTROSTATICS 32. An electric charge 10 8 C is placed at the point (4m, 7m, 2m). At the point (1m, 3m, 2m), the electric (A) potential will be 18 V (B) field has no Y-component (C) field will be along Z-axis (D) potential will be 1.8 V 33. When a negative charge is released and moves in electric field, it moves towards a position of (A) lower electric potential and lower potential energy (B) lower electric potential and higher potential energy (C) higher electric potential and lower potential energy (D) higher electric potential and higher potential energy 34. Figure shows equi-potential surfaces for a two charge-system. At which of the labeled points will an electron have the highest potential energy? A B q +q D C (A) Point A (B) Point B (C) Point C (D) Point D 35. An infinite nonconducting sheet of charge has a surface charge density of 10 7 C/m2. The separation between two equipotential surfaces near the sheet whose potential differ by 5V is (A) 0.88 cm (B) 0.88 mm (C) 0.88 m (D) 5 10 7 m 36. In a regular polygon of n sides, each corner is at a distance r from the centre. Identical charges are placed at (n 1) corners. At the centre, the intensity is E and the potential is V. The ratio V/E has magnitude (A) r n (B) r(n 1) (C) (n 1)/r (D) r(n 1)/n 37. A non-conducting ring of radius 0.5 m carries a total charge of 1.11 10 10 C distributed non-uniformly on its circumference producing an electric field every where in space. The value of the line integral (A) + 2 =0 = E d (B) 1 ( = 0 being centre of the ring) in volts is: (C) 2 (D) zero 38. A charge of 3 coulomb experiences a force of 3000 N when placed in a uniform electric field. The potential difference between two points separated by a distance of 1 cm along the field lines is (A) 10 V (B) 100 V (C) 1000 V (D) 9000 V 39. Uniform electric field of magnitude 100 V/m in space is directed along the line y = x + 3. Find the potential difference between point A (3, 1) & B (1, 3). (A) 100 V PHYSICS WALLAH (B) 200 2 V (C) 200 V (D) 0 V 70 ELECTROSTATICS 40. Variation of electrostatic potential along x-direction is shown in the graph. The correct statement about electric field is v A C B x (A) x component at point B is maximum (B) x component at point A is towards positive x-axis. (C) x component at point C is along negative x-axis (D) x component at point C is along positive x-axis 41. Points A, B, C, D, P and Q are shown in a region of uniform electric field. The potentials at some of the points are V(A) = 2 V & V(P) = V(B) = V(D) = 5 V. The electric field in the region is C 0.2m B P Q D A 0.2m (B) 15 2 V/m along PA (D) 5 V/m along PA (A) 10 V/m along PQ (C) 5 V/m along PC 42. 43. In the previous problem potential at point C is (A) 6.5 V (B) 8 V (C) 9.5 V (D) None of these The equation of an equipotential line in an electric field is y = 2x, then the electric field strength vector at (1, 2) may be (A) 4i + 8 j (B) 4i 8 j (C) 8i 4 j (D) 8i + 4 j 44. In a uniform electric field, the potential is 10 V at the origin of coordinates, and 8 V at each of the points(1, 0, 0), (0, 1, 0) and (0, 0, 1). The potential at the point (1, 1, 1) will be (A) 0 (B) 4 V (C) 8 V (D) 10 V 45. A uniform electric field having strength E is existing in x-y plane as shown in figure. Find the potential difference between origin O and point A(d, d, 0) E Y A( O ,0) d,d X Z (A) Ed(cos + sin ) (B) Ed(sin cos ) (C) PHYSICS WALLAH 2 Ed (D) 2 Ed 71 ELECTROSTATICS 46. In a certain region of space, the potential is given by : V = k[2x 2 y2 + z2]. The electric field at the point (1, 1, 1) has magnitude = (A) k 6 47. (B) 2k 6 (C) 2k 3 Two identical thin rings, each of radius R meter are coaxially placed at distance R meter apart. If Q1 and Q2 coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is (A) zero (C) 48. (D) 4k 3 (B) q 2 (Q1 + Q 2 ) (D) 4 0 R q(Q1 Q 2 ) ( 2 1 ) (4 2 0 R) q(Q1 Q 2 ) ( 2 + 1) (4 2 0 R) The diagram shows three infinitely long uniform line charges placed on the X, Y and Z axis. The work done in moving a unit positive charge from (1, 1, 1) to (0, 1, 1) is equal to Y 3 X 2 Z (A) 49. ln 2 (B) 2 0 ln 2 (C) 0 3 ln 2 (D) None of these 2 0 Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to : Q +q (A) 50. q (B) 1+ 2 2q 2+ (C) 2q (D) +q 2 Four equal charges +q are placed at four corners of a square with its centre at origin and lying in yz plane. The electrostatic potential energy of a fifth charge +q' varies on x-axis as: U U (A) U (B) x 51. +q x U (C) x x (D) x x x x Two positively charged particles X and Y are initially far away from each other and at rest. X begins to move towards Y with some initial velocity. The total momentum and energy of the system are p and E. (A) If Y is fixed, both p and E are conserved. (B) If Y is fixed, E is conserved, but not p. (C) If both are free to move, p is conserved but not E. (D) If both are free, E is conserved, but not p. PHYSICS WALLAH 72 ELECTROSTATICS 52. Two particles X and Y, of equal mass and with unequal positive charges, are free to move and are initially far away from each other. With Y at rest, X begins to move towards it with initial velocity u. After along time, finally (A) X will stop, Y will move with velocity u. (B) X and Y will both move with velocities u/2 each. (C) X will stop, Y will move with velocity < u. (D) both will move with velocities < u/2. 53. In space of horizontal electric field (E = (mg)/q) exist as shown in figure and a mass m attached at the end of a light rod. If mass m is released from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position mg E= q 45 m (A) 54. (B) 2g (C) 3g (D) 5g Two identical particles of mass m carry a charge Q each. Initially one is at rest in free space and the other is projected directly towards first particle from a large distance with speed v. The closed distance of approach be (A) 55. g +q Q2 1 4 0 mv (B) 1 4Q 2 (C) 4 0 mv 2 1 2Q 2 4 0 mv 2 (D) 1 3Q 2 4 0 mv 2 The diagram shows a small bead of mass m carrying charge q. The bead can freely move on the frictionless ring placed in horizontal plane. In the plane of the ring a charge +Q has also been fixed as shown. The potential at the point P due to +Q is V. The velocity with which the bead should projected from the point P, so that it can complete a circle, should be greater than g +Q a (A) 56. 6qV m (B) qV m 4a (C) P 3qV m (D) None A charged particle of charge Q is held fixed and another charged particle of mass m and charge q (of the same sign) is released from a distance r. The impulse of the force exerted by the external agent on the fixed charge by the time distance between Q and q becomes 2r is (A) Qq 4 0 mr PHYSICS WALLAH (B) Qqm 4 0 r (C) Qqm 0 r (D) Qqm 2 0 r 73 ELECTROSTATICS 57. A solid sphere of radius R is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the centre? (A) R (B) R/2 (C) R / 3 (D) 2 R 58. A small bullet of mass m and charge q is fired towards a solid uniformly charged large sphere of radius R and total charge +q. It enters the surface of sphere with speed u along a diameter of the sphere. Find the minimum speed u so that it can penetrate through the sphere. (Neglect all resistive forces acting on the bullet) (A) q 2 0 mR (B) q (C) 4 0 mR q 8 0 mR (D) 3q 4 0 mR 59. n small drops of same size are charged to V volts each. If they coalesce to form a single large drop, then its potential will be (A) V/n (B) Vn (C) Vn1/3 (D) Vn2/3 60. Four charges are placed each at a distance a from origin. The dipole moment of configuration is y 3q 2q 2q x q (A) 2q aj 61. (C) 2aq[i + j] (D) None The dipole moment of a system of charge +q distributed uniformly on an arc of radius R subtending an angle /2 at its centre where another charge q is placed is : (A) 62. (B) 3q aj 2 2qR (B) 2qR (C) qR (D) 2qR Figure shows the electric field lines around an electric dipole. Which of the arrows best represents the electric field at point P ? P (A) PHYSICS WALLAH (B) (C) (D) 74 ELECTROSTATICS 63. Point P lies on the axis of a dipole. If the dipole is rotated by 90 anti clock wise, the electric field vector E at P will rotate by (A) 90 clockwise (B) 180 clockwise (C) 90 anti-clockwise (D) it depends on which side of the dipole the point P is located. 64. A and B are two points on the axis and the perpendicular bisector respectively of an electric dipole. A and B are far away from the dipole and at equal distance from it. If the field at A and B are E A and E B then (A) E A = E B (B) E A = 2E B (C) E A = 2E B 1 (D) | E B |= 65. 2 | E A | and E B is perpendicular to E A A large sheet carries uniform surface charge density . A rod of length has a linear charge density on one half and on the second half. The rod is hinged at mid point O and makes an angle with the normal to the sheet. The torque experienced by the rod is + + + + + + + + + (A) 0 66. (B) 2 0 + 2 sin (C) 0 2 sin (D) 2 0 Two short electric dipoles are placed as shown. The energy of electric interaction between these dipoles will be P1 r P2 (A) 67. 2kP1P2 cos r 3 (B) 2kP1P2 cos r 3 (C) 2kP1P2 sin r 3 (D) 4kP1P2 cos r3 An uncharged sphere of metal placed inside a charged parallel plate capacitor. A charged parallel plate capacitor consists of two large parallel oppositely and uniformly charged plates. Which of the following figure correctly represents electric lines of force? (A) PHYSICS WALLAH (B) (C) (D) 75 ELECTROSTATICS 68. A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path (s) shown in figure as : 1 2 3 4 (A) 1 69. 1 2 3 4 (B) 2 (C) 3 (D) 4 An ellipsoidal cavity is carved within a perfect conductor. A positive charge q is placed at the center of the cavity. The points A & B are on the cavity surface as shown in the figure. Then A q B (A) electric field near A in the cavity = electric field near B in the cavity (B) charge density at A = charge density at B (C) potential at A = potential at B (D) total electric field flux through the surface of the cavity is q/ 70. A positive charge q is placed in a spherical cavity made in a positively charged sphere. Position vector of the centre of cavity relative to centre of the sphere is q is (A) in the direction parallel to vector . Force on the charge . (B) in the direction perpendicular to vector . (C) in a direction which depends on the magnitude of charge density in sphere. (D) direction can not be determined from the given information. 71. A conducting sphere of radius r has a charge. Then (A) The charge is uniformly distributed over its surface, if there is an external electric field. (B) Distribution of charge over its surface will be non uniform if no external electric field exists in space. (C) Electric field strength inside the sphere will be equal to zero only when no external electric field exists (D) Potential at every point of the sphere must be same 72. Two identical small conducting spheres, having charges of opposite sign, attract each other with a force of 0.108 N when separated by 0.5 m. The spheres are connected by a conducting wire, which is then removed, and thereafter, they repel each other with a force of 0.036 N. The initial charges on the spheres are (A) 5 10 6 C and 15 10 6 C (B) 1.0 10 6 C and 3.0 10 6 C (C) 2.0 10 6 C and 6.0 10 6 C (D) 0.5 10 6 C and 1.5 10 6 C PHYSICS WALLAH 76 ELECTROSTATICS 73. Two spherical, nonconducting, and very thin shells of uniformly distributed positive charge Q and radius d are located a distance 10d from each other. A positive point charge q is placed inside one of the shells at a distance d/2 from the center, on the line connecting the centers of the two shells, as shown in the figure. What is the net force on the charge q? Q Q d d/2 10 d (A) (C) qQ 361 0 d 2 to the left (B) 2 to the left (D) 362qQ 361 0 d qQ 361 0 d 2 360qQ 361 0 d 2 to the right to the right 74. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. The potential at the centre of the sphere is (A) 0 V (B) 10 V (C) same as at point 5 cm away from the surface out side sphere. (D) same as a point 25 cm away from the surface. 75. Two large metal plates each of area A carry charges +q and q and face each other as shown in the figure. The plates are separated by a small distance d. Charges appearing on surfaces A, B, C, D and electric field in the region between the plates are A 76. B C D (A) q q q q q , , , and 0A 2 2 2 2 (B) zero, q, q, zero and (C) q q q q 2q , , , and 0A 2 2 2 2 (D) zero, q, q and q 0A 2q 0A Two large metal plates each of area A carry charges +2q and q and face each other as shown in the figure. The plates are separated by a small distance d. Charges appearing on surfaces A, B, C, D are A (A) 0.5q, 1.5q, 1.5q, and +0.5q (C) zero, 2q, q and zero PHYSICS WALLAH B C D (B) q, q, q and zero (D) q, q, zero and q. 77 ELECTROSTATICS 77. In the given figure if the electric potential of the inner metal sphere is 10 V & that of the outer shell is 5 V, then the potential at the centre will be : a b (A) 10 V 78. (B) 5 V (C) 15 V (D) 0 V Three concentric metallic spherical shells A, B and C of radii a, b and c (a < b < c) have surface charge densities , + , and respectively. The potential of shell A is : (A) ( / 0 ) [a+b c] (B) ( / 0 ) [ a b + c ] (C) ( / 0 ) [b a c] (D) None Comprehension for Q. no. 79 and 80 A spherical shell with an inner radius a and an outer radius b is made of conducting material. A point charge +Q is placed at the centre of the spherical shell and a total charge q is placed on the shell. 79. Charge q is distributed on the surfaces as b Q a q (A) Q on the inner surface, q on outer surface (B) Q on the inner surface, q + Q on the outer surface (C) +Q on the inner surface, q Q on the outer surface (D) The charge q is spread uniformly between the inner and outer surface. 80. Assume that the electrostatic potential is zero at an infinite distance from the spherical shell. The electrostatic potential at a distance R(a < R < b) from the centre of the shell is (where K = (A) 0 PHYSICS WALLAH (B) KQ a (C) K Q q R (D) K 1 4 0 ) Q q b 78 ELECTROSTATICS EXERCISE #4 1. There are four concentric metallic shells A, B, C and D of radii a, 2a, 3a and 4a respectively. Shells B and D are given charges +q and q respectively. Shell C is now earthed. The potential difference VA VC is Kq Kq Kq Kq (A) (B) (C) (D) 2a 3a 4a 6a 2. Two fixed charges 4Q (positive) and Q (negative) are located at A and B, the distance AB being 3 m. Q +4Q A 3m B (A) The point P where the resultant field due to both is zero is on AB outside AB. (B) The point P where the resultant field due to both is zero is on AB inside AB. (C) If a positive charge is placed at P and displaced slightly along AB it will execute oscillations. (D) If a negative charge is placed at P and displaced slightly along AB it will execute oscillations. 3. Two point charges Q and Q/4 are separated by a distance x. Then Q/4 Q x (A) potential is zero at a point on the axis which is x/3 on the right side of the charge Q/4 (B) potential is zero at a point on the axis which is x/5 on the left side of the charge Q/4 (C) electric field is zero at a point on the axis which is at a distance x on the right side of the charge Q/4 (D) there exist two points on the axis where electric field is zero. 4. Three point charges Q, 4 Q and 16 Q are placed on a straight line 9 cm long. Charges are placed in such a way that the system has minimum potential energy. Then (A) 4 Q and 16 Q must be at the ends and Q at a distance of 3 cm from the 16 Q. (B) 4 Q and 16 Q must be at the ends and Q at a distance of 6 cm from the 16 Q. (C) Electric field at the position of Q is zero. (D) Electric field at the position of Q is 5. Q 4 0 Two infinite sheets of uniform charge density + and are parallel to each other as shown in the figure. Electric field at the + + + + + + + + + + (A) points to the left or to the right of the sheets is zero. (B) midpoint between the sheets is zero. (C) midpoint of the sheets is / 0 and is directed towards right. (D) midpoint of the sheet is 2 / 0 and is directed towards right. PHYSICS WALLAH 79 ELECTROSTATICS 6. Potential at a point A is 3 volt and at a point B is 7 volt, an electron is moving towards A from B. (A) It must have some K.E. at B to reach A (B) It need not have any K.E. at B to reach A (C) to reach A it must have more than or equal to 4 eV K. E. at B. (D) when it will reach A, it will have K.E. more than or at least equal to 4 eV if it was released from rest at B. 7. The figure shows a nonconducting ring which has positive and negative chargen on uniformly distributed on it such that the total charge is zero. Which of the following statements is true? O Axis (A) The potential at all the points on the axis will be zero. (B) The electric field at all the points on the axis will be zero. (C) The direction of electric field at all points on the axis will be along the axis. (D) If the ring is placed inside a uniform external electric field then net torque and force acting on the ring would be zero. 8. At distance of 5cm and 10cm outwards from the surface of a uniformly charged solid sphere, the potentials are 100V and 75V respectively. Then (A) potential at its surface is 150V. (B) the charge on the sphere is (5/3) 10 10C. (C) the electric field on the surface is 1500 V/m. (D) the electric potential at its centre is 225V. 9. Four identical charges are placed at the points (1, 0, 0), (0, 1, 0), ( 1, 0, 0) and (0, 1, 0). (A) The potential at the origin is zero. (B) The field at the origin is zero. (C) The potential at all points on the z-axis, other than the origin, is zero. (D) The field at all points on the z-axis, other than the origin acts along the z-axis. 10. A particle of charge 1 C & mass 1 gm moving with a velocity of 4 m/s is subjected to a uniform electric field of magnitude 300 V/m for 10 sec. Then it's final speed cannot be: (A) 0.5 m/s (B) 4 m/s (C) 3 m/s (D) 6 m/s 11. A proton and a deuteron are initially at rest and are accelerated through the same potential difference. Which of the following is true concerning the final properties of the two particles ? (A) They have different speeds (B) They have same momentum (C) They have same kinetic energy (D) none of these PHYSICS WALLAH 80 ELECTROSTATICS 12. Particle A having positive charge is moving directly head-on towards initially stationary positively charged particle B. At the instant when A and B are closest together. (A) the momenta of A and B must be equal (B) the velocities of A and B must be equal (C) B would have gained less kinetic energy than A would have lost. (D) B would have gained the same momentum as A would have lost 13. An electric dipole moment p = (2.0 i + 3.0 j) Cm is placed in a uniform electric field 105NC 1 E = (3.0 i + 2.0k) Nm. (A) The torque that E exerts on p is (0.6 i 0.4 j 0.9k) (B) The potential energy of the dipole is 0.6 J. (C) The potential energy of the dipole is 0.6 J. (D) If the dipole is rotated in the electric field, the maximum potential energy of the dipole is 1.3 J. 14. Which of the following is true for the figure showing electric lines of force? (E is electrical field, V is potential) B A (A) EA > EB 15. 16. (C) VA > VB (D) VB > BA If we use permittivity , resistance R, gravitational constant G and voltage V as fundamental physical quantities, then (A) [angular displacement] = 0R0G0V0 (B) [Velocity] = 1R 1G0V0 (C) [dipole moment] = 1R0G0V1 (D) [force] = 1R0G0V2 Units of electric flux are (A) 17. (B) EB > EA N m2 Coul 2 (B) N Coul m 2 2 (C) volt - m (D) volt-m3 Which of the following statements are correct? (A) Electric field calculated by Gauss law is the field due to only those charges which are enclosed inside the Gaussian surface. (B) Gauss law is applicable only when there is a symmetrical distribution of charges. (C) Electric flux through a closed surface will depends only on charges enclosed within that surface only. (D) None of these PHYSICS WALLAH 81 ELECTROSTATICS 18. Mark the correct options: (A) Gauss s law is valid only for uniform charge distributions. (B) Gauss s law is valid only for charges placed in vacuum. (C) The electric field calculated by Gauss s law is the field due to all the charges. (D) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface. 19. A hollow closed conductor of irregular shape is given some charge. Which of the following statements are correct? (A) The entire charge will appear on its outer surface. (B) All points on the conductor will have the same potential. (C) All points on its surface will have the same charge density. (D) All points near its surface and outside it will have the same electric intensity. 20. Charges Q1 and Q2 lies inside and outside respectively of a closed surface S. Let E be the field at any point on S and be the flux of E over S. (A) If Q1 changes, both E and will change. (B) If Q2 changes, E will change but will not change. (C) If Q1 = 0 and Q2 0 then E 0 but = 0. (D) If Q1 0 and Q2 = 0 then E = 0 but 0. 21. Three points charges are placed at the corners of an equilateral triangle of side L as shown in the figure. 2q L +q L L +q (A) The potential at the centroid of the triangle is zero. (B) The electric field at the centroid of the triangle is zero. 22. (C) The dipole moment of the system is 2 qL (D) The dipole moment of the system is 3 qL An electric dipole is placed at the centre of a sphere. Mark the correct answer (A) The flux of the electric field through the sphere is zero (B) The electric field is zero at every point of the sphere. (C) The electric potential is zero everywhere on the sphere. (D) The electric potential is zero on a circle on the surface. PHYSICS WALLAH 82 ELECTROSTATICS 23. For the situation shown in the figure below (assume r >> length of dipole) mark out the correct statement(s). p (Small dipole) Q r (A) Force acting on the dipole is zero. pQ (B) Force acting on the dipole is approximately (C) Torque acting on the dipole is (D) Torque acting on the dipole is pQ 4 0 r 3 and is acting upward in clockwise direction. 4 0 r 2 pQ in anti-clockwise direction 4 0 r 2 24. A particle of mass m and charge q is thrown in a region where uniform gravitational field and electric field are present. The path of particle (A) may be a straight line (B) may be a circle (C) may be a parabola (D) may be a hyperbola 25. For the situation shown in the figure below, mark out the correct statement(s) B q d R Hollow neutral conductor q (A) Potential of the conductor is (B) Potential of the conductor is 4 0 (d + R ) q 4 0 d (C) Potential of the conductor can t be determined as nature of distribution of induced charges is not known (D) Potential at point B due to induced charges is 26. qR 4 0 (d + R )d Two large thin conducting plates with small gap in between are placed in a uniform electric field E (perpendicular to the plates). Area of each plate is A and charges +Q and Q are given to these plates as shown in the figure. If points R, S and T as shown in the figure are three points in space, then the Q +Q R S T E (A) field at point R is E (B) field at point S is E Q (C) field at point T is E + 0A Q (D) field at point S is E + A 0 PHYSICS WALLAH 83 ELECTROSTATICS 27. In the shown figure the conductor is uncharged and a charge q is placed inside a spherical cavity at a distance a from its centre (C). Point P and charge +Q are as shown. a, b, c, d are known. +Q P d c C q a Column-I b Column-II (A) Electric field due to induced charges on (P) zero (Q) non-zero (R) value can be stated with the inner surface of cavity at point P (B) Electric potential due to charges on the inner surface of cavity and q at P (C) Electric field due to induced charges on the outer surface of conductor and Q at C (D) Electric potential due to induced charges the given data. (S) on the inner surface of cavity at C 28. value cannot be stated from the given data Column I shows graphs of electric potential V versus x and y in a certain region for four situations. Column II shows the range of angle which the electric field vector makes with positive x-direction Column I : V versus x, V versus y 45 (A) 0 300 135 0 (B) (C) Column II : Range of angle 0 0 60 45 30 (P) 00 < 45 (Q) 450 < 90 (R) 900 < 135 0 (D) PHYSICS WALLAH 0 60 (S) 1350 < 180 84 ELECTROSTATICS Paragraph for Question No. 29 to 31 Four metallic plates are placed as shown in the figure. Plate 2 is given a charge Q whereas all other plates are uncharged. Plates 1 and 4 are joined together. The area of each plate is same. 1 2 3 4 Q d 29. (C) 3Q/4 (B) +Q/4 (D) Q/2 The charge appearing on right side of plate 4 is (B) Q/4 (A) zero 31. d The charge appearing on the right side of plate 3 is (A) zero 30. 2d (C) 3Q/4 (D) +Q/2 The potential difference between plates 1 and 2 is (A) 3 Qd (B) 2 0A Qd (C) 0A 3 Qd (D) 4 0A 3Qd 0A Paragraph for Question No. 32 to 34 The sketch below shows cross sections of equipotential surfaces between two charged conductors that are shown in solid black. Some points on the equipotential surfaces near the conductors are marked as A, B, C,...... The arrangement lies in air. [Take = 8.85 10 12 C2/N m2] 0.3m Large conducting plate E Solid conducting sphere A B C D 30V 20V 10V 32. 33. 34. 10V 20V 30V 40V Surface charge density of the plate is equal to (A) 8.85 10 10 C/m2 (B) 8.85 10 10 C/m2 (C) 17.7 10 10 C/m2 (D) 17.7 10 10 C/m2 A positive charge is placed at B. When it is released (A) no force will be exerted on it (B) it will move towards A (C) it will move towards C (D) it will move towards E How much work is required to slowly move a 1 C charge from E to D ? (A) 2 10 5 J PHYSICS WALLAH (B) 2 10 5 J (C) 4 10 5 J (D) 4 10 5 J 85 ELECTROSTATICS ASSERTION AND REASON 35. Statement-1 : A positive point charge initially at rest in a uniform electric field starts moving along electric lines of forces. (Neglect all other forces except electric forces) Statement-2 : Electric lines of force represents path of charged particle which is released from rest in it. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. 36. Statement-1 : The electric potential and the electric field intensity at the centre of a square having four fixed point charges at their vertices as shown in figure are zero. Statement-2 : If electric potential at a point is zero then the magnitudeof electric field at that point must be zero. +q q q +q (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. 37. Statement-1 : In a given arrangement of charges, an extra charge is placed outside the Gaussian surface. In the Gauss Theorem E ds = Q in 0 : value of Qin remains unchanged whereas electric field E at the Gaussian surface is changed. Statement-2 : Electric field E at any point on the Gaussian surface is due to inside charge only. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. PHYSICS WALLAH 86 ELECTROSTATICS 38. When two charged concentric spherical conductors have electric potential V1 and V2 respectively. Statement-1 : The potential at centre is V1 + V2. Statement-2 : Potential is scalar quantity. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. 39. Statement-1 : A point charge q is placed inside a cavity of conductor as shown. Another point charge Q is placed outside the conductor as shown. Now as the point charge Q is pushed away from conductor, the potential difference (VA VB) between two points A and B within the cavity of conductor remains constant. Statement-2 : The electric field due to charges on outer surface of conductor and outside the conductor is zero at all points inside the conductor. A q B Q (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. PHYSICS WALLAH 87 ELECTROSTATICS EXERCISE #5 (JM) 1. An electric charge 10 3 C is placed at the origin (0,0) of X Y co-ordinate system. Two points A and B are situated at ( 2, 2 ) and (2,0) respectively. The potential difference between the points A and B will be (1) 9 volt (2) zero 2. (3) 2 volt [AIEEE-2007, 3/120] (4) 4.5 volt Charges are placed on the vertices of a square as shown. Let E be the electric field and V the potential at the centre. If the charges on A and B are interchanged with those on D and C respectively, then [AIEEE-2007, 3/120] (1) E remains unchanged, V changes (2) Both E and V change (3) E and V remain unchanged (4) E changes, V remains unchanged 3. The potential at a point x (measured in m) due to some charges situated on the x-axis is given by V(x) = 20/(x2 4) volts. The electric field E at x = 4 m is given by : [AIEEE-2007, 3/120] (1) 5/3 volt/ m and in the ve x direction (2) 5/3 volt/ m and in the +ve x direction (3) 10/9 volt/ m and in the ve x direction (4) 10/9 volt/ m and in the +ve x direction 4. A thin spherical shell of radius R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field E (r) produced by the shell in the range 0 < r < , where r is the distance from the centre of the shell? [AIEEE-2008, 3/105] (1) (2) (3) (4) 5. Two points P and Q are maintained at the potentials of 10 V and 4 V respectively. The work done in moving 100 electrons from P to Q is : [AIEEE-2009, 4/144] 17 16 16 (1) 9.60 10 J (2) 2.24 10 J (3) 2.24 10 J (4) 9.60 10 17 J 6. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then Q/q equals: [AIEEE-2009, 4/144] (1) 1 (2) 1 (3) 1 2 PHYSICS WALLAH (4) 2 2 88 ELECTROSTATICS 7. Statement 1 : For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q. [AIEEE-2009, 6/144] Statement 2 : The net work done by a conservative force on an object moving along a closed loop is zero. (1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (2) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is false. 8. Let (r) = r be the charge density distribution for a solid sphere of radius R and total R4 charge Q. For a point P inside the sphere at distance r1 from the centre of sphere, the magnitude of electric field is : [AIEEE-2009, 4/144] (1) 9. Q (2) 4 0 r12 Qr12 (3) 4 0 R 4 Qr12 (4) 0 3 0 R 4 A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field at the centre O is : [AIEEE-2010, 4/144] (1) 10. Q q 4 0 r 2 2 j (2) q 4 0 r 2 2 j (3) q 2 0 r 2 2 j (4) q 2 0 r 2 2 j Let there be a spherically symmetric charge distribution with charge density varying as 5 r (r) = 0 upto r = R, and (r) = 0 for r > R, where r is the distance from the origin. 4 R The electric field at a distance r (r < R) from the origin is given by (1) 4 0 r 5 r 3 0 3 R (2) 0 r 5 r 4 0 3 R (3) 4 0 r 5 r 3 0 4 R [AIEEE-2010, 4/144] (4) 0 r 5 r 3 0 4 R 11. Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30 with each other. When suspended in a liquid of density 0.8 g cm 3, the angle remains the same. If density of the material of the sphere is 1.6 g cm 3, the dielectric constant of the liquid is [AIEEE-2010, 8/144] (1) 4 (2) 3 (3) 2 (4) 1 12. The electrostatic potential inside a charged spherical ball is given by = ar2 + b where r is the distance from the centre; a,b are constants. Then the charge density inside the ball is : [AIEEE - 2011, 4/120, 1] (1) 24 a 0r (2) 6 a 0r (3) 24 a 0 (4) 6 a 0 PHYSICS WALLAH 89 ELECTROSTATICS 13. Two positive charges of magnitude q are placed at the ends of a side (side 1) of a square of side 2a . Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is : [AIEEE 2011, 11 May; 4, 1] (1) zero (3) 14. 2qQ 2 1 4 0 a 5 1 (2) 2qQ 1 1 + 4 0 a 5 (4) 2qQ 1 1 4 0 a 5 1 1 In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as function of distance from the centre. The graph which would correspond to the above will be : [AIEEE 2012 ; 4/120, 1] (1) 15. (2) (3) (4) This question has statement-1 and statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. [AIEEE 2012 ; 4/120, 1] An insulating solid sphere of radius R has a uniformly positive charge density . As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinite is zero. Statement-1 : When a charge q is taken from the centre to the surface of the sphere its q potential energy changes by . 3 0 Statement-2 : The electric field at a distance r (r < R) from the centre of the sphere is r 3 0 (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of statement-1. (2) Statement-1 is true Statement-2 is false. (3) Statement-1 is false Statement-2 is true. (4) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement-1. 16. Two charges, each equal to q, are kept at x = a and x = a on the x-axis. A particle of mass m and charge q q = 0 2 is placed at the origin. If charge q0 is given a small displacement (y <<a) along the y-axis, the net force acting on the particle is proportional to: (1) y PHYSICS WALLAH (2) y (3) 1 y [JEE-Mains 2013, 4/120] (4) 1 y 90 ELECTROSTATICS 17. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at distance L from the end A is : [JEE-Mains 2013, 4/120] O (1) Q 8 0 L (2) L 3Q 4 0 L A B L (3) Q 4 0 L ln 2 (4) Q ln 2 4 0 L 18. Assume that an electric field E = 30 x 2i exists in space. Then the potential difference VA VO, where VO is the potential at the origin and VA the potential at x = 2 m is : [JEE- Main - 2014] (1) 120 J (2) 120 J (3) 80 J (4) 80 J 19. A long cylindrical shell carries positive surface charge in the upper half and negative surface charge in the lower half. The electric field lines around the cylinder will look like figure given in ; (figures are schematic and not drawn to scale) [JEE MAIN - 2015] (1) (3) (2) (4) 20. A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ) on 3V0 5V0 3V0 V , , its surface. For this sphere the equipotential surfaces with potentials and 0 2 4 4 4 have radius R1, R2, R3 and R4 respectively. Then [JEE MAIN - 2015] (1) R1 1and (R2 R1) > (R4 R3) (2) R1 = 0 and R2 < (R4 R3) (3) 2R < R4 (4) R1 = 0 and R2 > (R4 R3) 21. The region between two concentric spheres of radii a and b , respectively (see figure), has volume charge density = A r , where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is : [JEE MAIN - 2016] Q 2Q 2Q Q (1) (2) (3) (4) 2 2 2 2 2 a 2 a 2 2 ( b a ) (a b ) 22. An electric dipole has a fixed dipole moment p , which makes angle with respect to x-axis, When subjected to an electric field E1 = Ei , it experiences a torque T1 = k . When subjected to another electric field E 2 = 3E1 j it experiences a torque T2 = T1 . The angle is : [JEE MAIN - 2017] (1) 30 PHYSICS WALLAH (2) 45 (3) 60 (4) 90 91 ELECTROSTATICS EXERCISE #6 (JA) 1. 2. PREVIOUS YEAR IIT JEE ADVANCED QUESTIONS Two equal point charges are fixed at x = a and x = +a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to [IIT JEE-2002 (Scr)] 2 3 (A) x (B) x (C) x (D) 1/x A point charge q is placed at a point inside a hollow conducting sphere. Which of the following electric force pattern is correct ? [IIT JEE-2003 (scr)] (A) 3. (B) (C) (D) Charges +q and q are located at the corners of a cube of side a as shown in the figure. Find the work done to separate the charges to infinite distance. [IIT JEE-2003] q +q +q q +q q q +q 4. A charge +Q is fixed at the origin of the co-ordinate system while a small electric dipole of dipole-moment p pointing away from the charge along the x-axis is set free from a point far away from the origin. [IIT JEE-2003] (a) calculate the K.E. of the dipole when it reaches to a point (d, 0) (b) calculate the force on the charge +Q at this moment. 5. Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be due to [IIT JEE-2004 (SCR)] q2 +q1 q1 (A) q2 (C) all the charges 6. (B) only the positive charges (D) +q1 and q1 Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at O is double the electric field when only one positive charge of same magnitude is placed at R. Which of the following arrangements of charges is possible for P, Q, R, S, T and U respectively? IIT JEE-2004 (SCR)] P O U T (A) +, , +, , , + PHYSICS WALLAH (B) +, , +, , +, Q R S (C) +, +, , +, , (D) , +, +, , +, 92 ELECTROSTATICS 7. Two uniformly charged infinitely large plane sheets S1 and S2 are held in air parallel to each other with separation d between them. The sheets have charge distributions per unit area 1 and 2 (Cm 2), respectively, with 1 > 2. Find the work done by the electric field on a point charge Q that moves from S1 towards S2 along a line of length a (a > d) making an angle of /4 with the normal to the sheets. Assume that the charge Q does not affect the charge distributions of the sheets. 8. [IIT JEE-2004] Three large parallel plates have uniform surface charge densities as shown in the figure. Find out electric field intensity at point P. [IIT JEE-2005 (Scr.)] k z=a P 2 z = a (A) 9. 4 k 0 (B) z = 2a 4 k 0 (C) 2 k 0 (D) Which of the following groups do not have same dimensions 2 k 0 [IIT JEE-2005 (Scr.)] (A) Young s modulus, pressure, stress (B) Work, heat, energy (C) Electromotive force, potential difference, voltage (D) Electric dipole moment, electric flux, electric field 10. A conducting liquid bubble of a and thickness t (t<< a) is charged to potential V. If the bubble collapses to a droplet, find the potential on the droplet. 11. [IIT JEE-2005] The electrostatic potential ( r ) of a spherical symmetric system, kept at origin, is shown in the adjacent figure, and given as r = q 4 0 r (r R 0 ) ; r = [IIT JEE-2006] q 4 0 R 0 (r R 0 ) Which of the following option(s) is/are correct? r R0 r (A) For spherical region r R0, total electrostatic energy stored is zero. (B) Within r = 2R0, total charge is q. (C) There will be no charge anywhere except at r = R0. (D) Electric field is discontinuous at r = R0. PHYSICS WALLAH 93 ELECTROSTATICS 12. A long hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. [IIT JEE-2007] (A) A potential difference appears between the two cylinders when a charge density is given to the inner cylinder. (B) A potential difference appears between the two cylinders when a charge density is given to the outer cylinder. (C) No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders. (D) No potential difference appears between the two cylinders when same charge density is given to both the cylinders. 13. Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. Then the net charge on the sphere is : [IIT JEE-2007] (A) negative and distributed uniformly over the surface of the sphere. (B) negative and appears only at the point on the sphere closest to the point charge. (C) negative and distributed non-uniformly over the entire surface of the sphere. (D) zero. 14. A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is : [IIT JEE-2007] 15. (A) zero every where (B) non zero and uniform (C) non uniform (D) zero only at its centre a a Positive and negative point charges of equal magnitude are kept at 0, 0, and 0, 0, , 2 2 respectively. The work done by the electric field when another positive point charge is moved from ( a, 0, 0) to (0, a, 0) is [IIT JEE-2007] (A) positive (B) negative (C) zero (D) depends on the path connecting the initial and final positions. PHYSICS WALLAH 94 ELECTROSTATICS 16. Consider a system of three charges q q , 3 3 and 2q 3 placed at points A, B and C, respectively, as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB = 60 . y B C 60 x O A [IIT JEE-2008] (A) The electric field at point O is q 8 0 R 2 directed along the negative x-axis. (B) The potential energy of the system is zero. (C) The magnitude of the force between the charges at C and B is (D) The potential at point O is q2 . 54 0 R 2 q 12 0 R Paragraph for Question Nos. 17 to 19 The nuclear charge (Ze) is non uniformly distributed within a nucleus of radius R. The charge density (r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial direction. (r) d a 17. R The electric field at r = R is : (A) independent of a (C) directly proportional to a2 r [IIT JEE-2008] (B) directly proportional to a (D) inversely proportional to a 18. For a = 0, the value d (maximum value of as shown in the figure) is : 3Ze 3Ze 4Ze Ze (A) (B) (C) (D) 3 3 3 R 3 R 3 R 3 4 R 19. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies : R 2R (A) a = 0 (B) a = (C) a = R (D) a= 2 3 20. A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at ( a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4 to x = 5a/4. Two point charges 7C and 3C are placed at (a/4, a/4, 0) and ( 3a/4, 3a/4, 0), respectively. Consider a cubical surface formed by six surfaces x = a/2, y = a/2, z = a/2. The electric flux through this cubical surface is : [IIT JEE -2009] y x (A) 2C 0 PHYSICS WALLAH (B) 2C 0 (C) 10C 0 (D) 12C 0 95 ELECTROSTATICS 21. Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1 : Q2 : Q3, is [IIT JEE -2009] (A) 1 : 2 : 3 (B) 1 : 3 : 5 (C) 1 : 4 : 9 (D) 1 : 8 : 18 22. Under the influence of the Coulomb field of charge +Q, a charge q is moving around it in an elliptical orbit. Find out the correct statement(s). [IIT JEE -2009] (A) The angular momentum of the charge q is constant. (B) The linear momentum of the charge q is constant. (C) The angular velocity of the charge q is constant. (D) The linear speed of the charge q is constant. 23. A solid sphere of radius R has a charge Q distributed in its volume with a charge density = Kra, where K and a are constants and r is the distance from its centre. If the electric field at R 1 r= is times that at r = R, find the value of a. [IIT JEE-2009] 2 8 24. A uniformly charged thin spherical shell of radius R carries uniform surface charge density of per unit area. It is made of two hemispherical shells, held together by pressing them with force F (see figure). F is proportional to [IIT JEE-2010] F (A) 25. 1 0 R 2 2 (B) 1 0 R 2 F (C) 1 2 0 R (D) 1 2 0 R 2 A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform 81 105 Vm 1. When the field is switched off, the drop is observed electric field of strength 7 to fall with terminal velocity 2 10 3 m/s. Given g = 9.8 m s 2, viscosity of the air = 1.8 10 5 Ns m 2 and the density of oil = 900 kg m 3, the magnitude of q is : [IIT JEE-2010] 19 19 19 (A) 1.6 10 C (B) 3.2 10 C (C) 4.8 10 C (D) 8.0 10 19 C 26. A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that : [IIT JEE-2010] (A) |Q1| > |Q2| (B) |Q1 | < |Q2| (C) at a finite distance to the left of Q1 the electric field is zero (D) at a finite distance to the right of Q2 the electric field is zero PHYSICS WALLAH 96 ELECTROSTATICS 27. Consider an electric field E = E0 x , where E0 is a constant. The flux through the shaded area (as shown in the figure) due to this field is : [IIT JEE-2011] z (a,0,a) (a,a,a) (0,0,0) (0,a,0) 0 x (A) 2E0a2 (B) y 2 E0a2 (C) E0a2 (D) E 0a 2 2 28. A spherical metal shell A of radius RA and a solid metal sphere B of radius RB (< RA) are kept far apart and each is given charge +Q . Now they are connected by a thin metal wire. Then [IIT JEE-2011] (A) E inside =0 A 29. (B) QA > QB (C) A RB = B RA surface surface (D) E on E on A B A wooden block performs SHM on a frictionless surface with frequency, 0. The block carries a charge +Q on its surface. If now a uniform electric field E is switched-on as shown, then the SHM of the block will be [IIT JEE-2011] E +Q (A) of the same frequency and with shifted mean position. (B) of the same frequency and with the same mean position. (C) of changed frequency and with shifted mean position. (D) of changed frequency and with the same mean position. 30. Which of the following statement(s) is/are correct? [IIT JEE-2011] (A) If the electric field due to a point charge varies as r 2.5 instead of r 2, then the Gauss law will still be valid. (B) The Gauss law can be used to calculate the field distribution around an electric dipole. (C) If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same. (D) The work done by the external force in moving a unit positive charge from point A at potential VA to point B at potential VB is (VB VA). PHYSICS WALLAH 97 ELECTROSTATICS 31. Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field E ( r ) and the electric potential V(r) with the distance r from the centre, is best represented by which graph? [IIT JEE-2012] (A) (B) (C) (D) 32. Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of potential difference X. A proton is released at rest midway between the two plates. It is found to move at 45 to the vertical JUST after release. Then X is nearly [IIT JEE-2012] (A) 1 10 5 V (B) 1 10 7 V (C) 1 10 9 V (D) 1 10 10 V 33. A cubical region of side a has its centre at the origin. It encloses three fixed point charges, q a a at 0, , 0 , +3q at (0, 0, 0) and q at 0, + , 0 , Choose the correct option(s) 4 4 [IIT JEE-2012] a (A) The net electric flux crossing the plane x = + plane x = is equal to the net electric flux crossing the 2 a 2 (B) The net electric flux crossing the plane y = + the plane y = a is more than the net electric flux crossing 2 a 2 (C) The net electric flux crossing the entire region is (D) The net electric flux crossing the plane z = + plane x = + PHYSICS WALLAH a 2 q 0 is equal to the net electric flux crossing the a 2 98 ELECTROSTATICS 34. An infinitely long solid cylinder of radius R has a uniform volume charge density . It has a spherical cavity of radius R/2 with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point P, which is at a distance 2R from the axis of the cylinder, is given by the expression 35. 23 R 16k 0 . The value of k is [IIT JEE-2012] Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown in the figure. Given that K = 1 q 4 0 L2 , which of the following statement (s) is (are) correct? [IIT JEE-2012] (A) The electric field at O is 6K along OD (B) The potential at O is zero (C) The potential at all points on the line PR is same (D) The potential at all points on the line ST is same. 36. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities 1 and 2 respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio can be (A) 4 37. 1 2 [IIT-JEE 2013] (B) 32 (C) 25 32 25 (D) 4 Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge densities + and , respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region : [IIT-JEE 2013] R1 R2 (A) the electrostatic field is zero (B) the electrostatic potential is constant (C) the electrostatic field is constant in magnitude (D) the electrostatic field has same direction PHYSICS WALLAH 99 ELECTROSTATICS 38. Let E1(r), E2 (r) and E3 (r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density , and an infinite plane with uniform surface charge density . If E1(r0) = E2 (r0) = E3 (r0) at a given distance r0, then 39. [IIT JEE -2014] (A) Q = 4 r02 (B) r0 = (C) E1 (r0 / 2) = 2E 2 (r0 / 2) (D) E 2 (r0 / 2) = 4E 3 (r0 / 2) 2 Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2, R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are E1, E2 and E3 respectively, then [IIT JEE -2014] P P R P R Q R 4Q 2Q R/2 2R Sphere 1 (A) E1 > E2 > E3 40. Sphere 2 (B) E3 > E1 > E2 Sphere 3 (C) E2 > E1 > E3 (D) E3 > E2 > E1 Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x axis at x = 2a, a, +a and +2a, respectively. A positive charge q is placed on the positive y axis at a distance b > 0. Four options of the signs of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match List I with List II and select the correct answer using the code given below the lists. [IIT JEE -2014] (+0, b) q Q1 Q2 ( 2a, 0) ( a, 0) Q3 Q4 (+a, 0) (+2a, 0) List I List II P. Q1, Q2, Q3 Q4 all positive 1. +x Q. Q1, Q2 positive; Q3, Q4 negative 2. x R. Q1, Q4 positive ; Q2, Q3 negative 3. +y S. Q1, Q3 positive; Q2, Q4 negative 4. y Code: (A) P-3, Q-1, R-4, S-2 (B) P-4, Q-2, R-3, S-1 (C) P-3, Q-1, R-2, S-4 (D) P-4, Q-2, R-1, S-3 PHYSICS WALLAH 100 ELECTROSTATICS 41. An infinitely long uniform line charge distribution of charge per unit length lies parallel to the y-axis in the y-z plane at z = 3 2 a (see figure). If the magnitude of the flux of the electric field through the rectangular surface ABCD lying in the x-y plane with its center at the origin is ( 0 = permittivity of free space), then the value of n is L n 0 [IIT JEE -2015] z L D C a 3a 2 y O A B x 42. The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density are kept parallel to each other. In their resulting electric field, point charges q and q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is(are) [IIT JEE -2015] x +q q x (A) Both charges execute simple harmonic motion. (B) Both charges will continue moving in the direction of their displacement. (C) Charge +q executes simple harmonic motion while charge q continues moving in the direction of its displacement. (D) Charge q executes simple harmonic motion while charge +q continues moving in the direction of its displacement. 43. Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this distribution, a spherical cavity of radius R2, centred at P with distance OP = = R1 R2 (see figure) is made. If the electric field inside the cavity at position r is E(r ) , then the correct statement(s) is(are) [IIT JEE -2015] R2 P R1 O (A) E is uniform, its magnitude is independent of R2 but its direction depends on r (B) E is uniform, its magnitude depends on R2 and its direction depends on r (C) E is uniform, its magnitude is independent of but its direction depends on (D) E is uniform and both its magnitude and direction depend on PHYSICS WALLAH 101 ELECTROSTATICS 44. A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct? [IIT JEE -2017] +Q R (A) The electric flux passing through the curved surface of the hemisphere is (B) Total flux through the curved and the flat surfaces is Q 1 1 2 0 2 Q 0 (C) The component of the electric field normal to the flat surface is constant over the surface (D) The circumference of the flat surface is an equipotential 45. The potential energy of a particle of mass m at a distance r from a fixed-point O is given by V(r) = kr2/2, where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true? [IIT JEE-2018] (A) v = 46. k 2m R (B) v = k m R (C) L = mkR 2 (D) L = mk 2 R2 An infinitely long thin non-conducting wire is parallel to the z-axis and carries a uniform line charge density . It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ subtends an angle 120 at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is 0. Which of the following statements is (are) true? [IIT JEE-2018] (A) The electric flux through the shell is 3R (B) The z-component of the electric field is zero at all the points on the surface of the shell (C) The electric flux through the shell is 2R (D) The electric field is normal to the surface of the shell at all points PHYSICS WALLAH 102 ELECTROSTATICS 47. A particle, of mass 10 3 kg and charge 1.0C , is initially at rest. At time t = 0 , the particle comes under the influence of an electric field E(t) = E 0 sin t , where E 0 = 1.0 NC 1 and = 103 rads 1 . Consider the effect of only the electrical force on the particle. Then the maximum speed, in ms 1 , attained by the particle at subsequent times is __________: [IIT JEE-2018] 48. The electric field E is measured at a point P(0, 0, d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d . List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II. [IIT JEE-2018] LIST-I LIST-II P. E is independent of d 1. A point charge Q at the origin Q. E 1 2. d A small dipole with point charges Q at (0, 0, l) and Q R. E 1 3. d2 at (0, 0, l) . Take 2l d An infinite line charge coincident with the x-axis, with uniform linear charge density S. E 1 4. d3 Two infinite wires carrying uniform linear charge density parallel to the x -axis. The one along (y = 0, z = l) has a charge density + and the one along (y = 0, z = l) has a charge density . Take 2l 5. d Infinite plane charge coincident with the xyplane with uniform surface charge density (A) P 5; Q 3, 4; R 1; S 2; (B) P 5; Q 3; R 1, 4; S 2; (C) P 5; Q 3; R 1, 2; S 4; (D) P 4; Q 2, 3; R 1; S 5; 49. A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential at its surface is V0 . A hole with a small area 4 R 2 ( 1) is made on the shell without affecting the rest of the shell. Which one of the following statements is correct? [IIT JEE-2019] (A) The potential at the center of the shell is reduced by 2 V0 (B) The magnitude of electric field at the center of the shell is reduced by V0 2R (C) The ratio of the potential at the center of the shell to that of the point at towards the hole will be 1 2 R from center 1 1 2 (D) The magnitude of electric field at a point, located on a line passing through the hole and shell's center, on a distance $2 R$ from the center of the spherical shell will be reduced by V0 2R PHYSICS WALLAH 103 ELECTROSTATICS 50. A charged shell of radius R carries a total charge Q. Given as the flux of electric field through a closed cylindrical surface of height h, radius r and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct? [ 0 is the permittivity of free space] [IIT JEE-2019] (A) If h > 2R and r > R then = Q/ 0 (B) If h < 8R/5 and r = 3R/5 then = 0 (C) If h > 2R and r = 3R/5 then = Q/5 0 (D) If h > 2R and r = 4R/5 then = Q/5 0 51. An electric dipole with dipole moment p9 ( + ) is held fixed at the origin O in the presence of 2 an uniform electric field of magnitude E 0 . If the potential is constant on a circle of radius R centered at the origin as shown in figure, then the correct statement(s) is/are: ( 0 is permittivity of free space. R >> dipole size) [IIT JEE-2019] 1/3 4 E 0 E 0 (A) R = p0 (B) Total electric field at point A is E A = 2E 0 ( + ) (C) Total electric field at point B is E B = 0 (D) The magnitude of total electric field on any two points of the circle will be same. 52. A uniform electric field, E = 400 3 yNC 1 is applied in a region. A charged particle of mass m carrying positive charge q is projected in this region with an initial speed of 2 10 106 ms 1 . This particle is aimed to hit a target T, which is 5 m away from its entry point into the field as shown schematically in the figure. Take q m = 1010 Ckg 1 . Then [IIT JEE-2020] (A) the particle will hit T if projected at an angle 45 from the horizontal (B) the particle will hit T if projected either at an angle 30 or 60 from the horizontal (C) time taken by the particle to hit T could be (D) time taken by the particle to hit T is PHYSICS WALLAH 5 3 5 6 s as well as 5 2 s s 104 ELECTROSTATICS 53. A circular disc of radius R carries surface charge density (r) = 0 1 r , where 0 is a R constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is 0 . Electric flux through another spherical surface of radius R 4 and concentric with the disc is . Then the ratio 0 is ______. [IIT JEE-2020] 54. A point charge q of mass m is suspended vertically by a string of length l. A point dipole of dipole moment p is now brought towards q from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is N (mgh), where g is the acceleration due to gravity, then the value of N is (Note that for three coplanar forces keeping a point mass in equilibrium, F sin is the same for all forces, where F is any one of the forces and is the angle between the other two forces) [IIT JEE-2020] 55. Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting, massless strings of same length. At equilibrium, the angle between the strings is . The spheres are now immersed in a dielectric liquid of density 800 kg m 3 and dielectric constant 21. If the angle between the strings remains the same after the immersion, then: [IIT JEE-2020] (A) electric force between the spheres remains unchanged (B) electric force between the spheres reduces (C) mass density of the spheres is 840 kg m 3 (D) the tension in the strings holding the spheres remains unchanged 56. Two point charges and + / 3 are placed in the xy-plane at the origin (0, 0) and a point (2, 0), respectively, as shown in the figure. This results in an equipotential circle of radius R and potential = 0 in the xy-plane with its center at (b, 0). All lengths are measured in meters. [IIT JEE-2021] (1) The value of is ___ meter. PHYSICS WALLAH (2) The value of is ___ meter. 105 ELECTROSTATICS 57. A charge q is surrounded by a closed surface consisting of an inverted cone of height h and base radius R, and a hemisphere of radius R as shown in the figure. The electric flux through the conical surface is 58. nq 6 (in SI units). The value of n is________: [IIT JEE-2022] 0 In the figure, the inner (shaded) region A represents a sphere of radius rA = 1 , within which the electrostatic charge density varies with the radial distance r from the center as A = kr , where k is positive. In the spherical shell B of outer radius rB , the electrostatic charge density varies as B = 2k r . Assume that dimensions are taken care of. All physical quantities are in their SI units. Which of the following statement(s) is(are) correct? 3 (A) If rB = (B) If rB = 2 3 2 [IIT JEE-2022] , then the electric field is zero everywhere outside B. , then the electric potential just outside B is k . 0 (C) If rB = 2 , then the total charge of the configuration is 15 k . (D) If rB = 5 2 , then the magnitude of the electric field just outside B is PHYSICS WALLAH 13 k . 0 106 ELECTROSTATICS 59. An electric dipole is formed by two charges + and located in -plane at (0,2) mm and (0, 2) mm, respectively, as shown in the figure. The electric potential at point P (100,100) mm due to the dipole is 0. The charges + and are then moved to the points ( 1,2) mm and (1, 2) mm, respectively. What is the value of electric potential at P due to the new dipole? [IIT JEE-2023] (A) V0/4 PHYSICS WALLAH (B) V0/2 (C) V0/ 2 (D) 3V0/4 107 ELECTROSTATICS ANSWER KEY EXERCISE #1 1. 2. a = (1 + 2) outside line segment AB and near to B, the equilibrium will be stable. 9 x0 x0 x (ii) 5. 9.30 3 11 9. F= 6. q2 6 10. 2 x O (i) 4 0 a E E O 4. q 3. E O 3 10 9 C, No. q 24 0 7. 1/3 11. a= R 8. 2 0 u 2 m 12. N = 11 q 3 1.8 105 sec 18. 20 ln 2 19. 21. q 2 tan 1 2 0 mg 22. 9V0 23. 26. 4 2y x (iv) 17. ( i 2 j) x0 3/ 2 14. (a) 7.08 10 8 (b) No 3 O (iii) 2 kP x x0 13. 25. E 15. 27. 3 16. kq 2 a (3 2 ) 2kQ 2 mR 7 1 qQ 2 4 0 mR 3 q1 1 1 Vr = ; a r b 4 0 r a q 1 1 b 29. (i) q 2 = q1 (ii) Vb = 1 ; r = b 4 0 b a a 1 q1 q 2 Vr = + ; r b 4 0 r r 30. 20. 5.86 m/s 24. 0 28. Q/3 1.125 q EXERCISE #2 1. zero q 2. 3. 2 0 m 4 Kq 2 8 = , Wsecond step = 0, Wtotal = 0 5. 5 r 3 4. Wfirst step 6. RE 0 i 10. 2KQq r R 3 mR r + 8 PHYSICS WALLAH 7. 4kq i R 2 1/ 2 11. 8. H2 = h1 + h2 g V 2 4 0 Ka Qq 2 0 L 9. 2 6 2 mr 0 e a 108 ELECTROSTATICS 12. (a) H = 4a a (b) U=mg 2 h 2 + a 2 h equilibrium at h = 3 3 U 2mga 3mga h a/ 3 13. kQr 4R 3r R (c) 3 R 14. v0 = 3 m/s.; K.E. at the origin = (27 10 6 ) 10 4 J approx 2.5 10 4 J 16. 5 17. 18. E dA = No 15. q in 0 E.A = Zero for equipotential E A E = Zero Volume inside ie equipotential 19. The total charge enclosed by a surface is zero, does not imply that the electric field everywhere on the surface is zero, it means only net flux is zero. If the electric field everywhere on a surface is zero, implies that net flux through the surface is zero, that means charge inside the surface is zero. 20. E O1 = 7 R 12 0 EXERCISE #3 1. 8. 15. 22. 29. 36. 43. 50. 57. 64. 71. 78. B B B C D B D B C C D C 2. 9. 16. 23. 30. 37. 44. 51. 58. 65. 72. 79. D D B D A A B B B B B B 3. 10. 17. 24. 31. 38. 45. 52. 59. 66. 73. 80. D C B A B A A A D B A D 4. 11. 18. 25. 32. 39. 46. 53. 60. 67. 74. D D C D A D B B A B B 5. 12. 19. 26. 33. 40. 47. 54. 61. 68. 75. B A C A C D B B A D B 6. 13. 20. 27. 34. 41. 48. 55. 62. 69. 76. B A C B B B B A B C A 7. 14. 21. 28. 35. 42. 49. 56. 63. 70. 77. A C D A B C B B A A A EXERCISE #4 1. 8. 15. 22. D ACD ABD AD 2. 9. 16. 23. PHYSICS WALLAH AD BD C BC 3. 10. 17. 24. ABC A C AC 4. 11. 18. 25. BC AC CD AD 5. 12. 19. 26. AC 6. BCD 13. AB 20. AD AC 7. ABD 14. ABC 21. A AD AD 109 ELECTROSTATICS 27. (A) QS; (B) PR; (C) PR; (D) QR 28. (A) S; (B) P; (C) R; (D) Q 29. B 30. D 31. C 32. A 36. C 37. C 38. D 39. A 33. B 34. D 35. C EXERCISE #5 (JM) 1. (2) 2. (4) 3. (4) 4. (4) 5. (3) 6. (4) 7. (1) 8. (2) 9. (3) 10. (2) 11. (3) 12. (4) 13. (4) 14. (3) 15. (3) 16. (1) 17. (4) 18. (3) 19. (4) 20. (2,3) 21. (4) 22. (3) EXERCISE #6 (JA) 1. B 2. 4. (a) K.E. = 5. C P A Q 4 0 d 2 ; (b) 3. QP 2 0 d 3 1 q2 4 0 a 4 [3 3 3 6 2 ] 6 along positive x-axis 6. D 7. ( 1 2 ) Qa a 10. V ' = V 3t 11. ABCD 12. 15. C 16. C 20. A 21. 25. D 30. 8. C 9. D A 13. D 14. B 17. A 18. B 19. C B 22. A 23. 2 24. A 26. AD 27. C 28. ABCD 29. A C or CD 31. D 32. C 33. ACD 34. 6 35. ABC 36. BD 37. CD 38. C 39. C 40. A 41. 6 42. C 43. D 44. AD 45. 50. 55. (B, C) (B, C, D) (A, C) 46. 51. 56. (A, B) (A, C) (R= 1.732) 47. 52. 57. (02) (B, C) (3) 48. 53. 58. (B) (b.4) (A) 49. 54. 59. (C) (02) (B) 2 2 0 1/3 PW Web/App - https://smart.link/7wwosivoicgd4 Library- https://smart.link/sdfez8ejd80if PHYSICS WALLAH 110

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