Electrostat-03-Subjective Solved

SOLVED PROBLEMS (SUBJECTIVE) 1. Three point charges q, 2q and 8q are to be placed on a 9 cm long straight line. Find the positions where the charges should be placed such that the potential energy of this system is minimum. In this situation, what is the electric field at the charge q due to the other two charges ? Solution: The maximum contribution may come from the charge 8q forming pairs with others. To reduce its effect, it should be placed at a corner and the smallest charge q in the middle. This arrangement shown in figure ensures that the charges in the strongest pair 2q, 8q are at the largest separation. The potential energy is 2q q 8q q2  2 16 8  U    . 40  x 9cm 9cm  x  This will be minimum if A  2  x 8 9cm  x is minimum. For this, dA   2  9cm  x2  0 …. (i) or, 9 cm – x = 2x or, x = 3 cm The electric field at the position of charge q is q  2 8      9cm  x2  = 0. From (i). 2. Four point charges +8  C, -1  C, -1  C and +8  C are fixed at the points - m, - m, + 3 / 2 m and + m respectively on the y-axis. A particle of mass 6 x 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. 1/(4   ) = 9 x 109 Nm2 / C2. Solution : in the figure q = 1  C = 10-6 C, and q = + 0.1  C = 10-7 C -4 y Q = 8  C = 8 × 10-6 C. B Let P be any point at a distance x from origin. Then A AP = CP = C D BP = DP = 27 / 2  x 2 Electric potential at point P will be: V  2KQ  2Kq , where BP AP K  1 4πε 0  9x109 Nm 2 / C 2   9  8  10 6 10 6  V  2  9.0  10     … (1)  27  x 2 3  x 2   2 2   Electric field at P is : dV   1  27 3 / 2  1  3   E    1.8 104  8     x2   2x 1    x2  2x dx E = 0 on x-axis where 8    1 2  2   2  2    27  3 / 2  x 2  3 / 2   x 2   2   2    27 43 / 2 3 / 2 2 1 3 / 2 2   x   2    x   2   27  x 2    3  x 2    4   2   2  x   5 m 2 The least value of kinetic energy of the particles of affinity should be enough to take the particles upto x   because At x   For x > and For x < E = 0 Fe = 0 Fe is repulsive (towards positive x-axis) 5 / 2 m ; Fe is attractive (towards negative x-axis) Now, from equation (1), potential at x = ,    8 1     V = 1.8 x 104  5 3 5     2  V = 2.7 x 104 volt. 2  2  Applying energy conservation at x =  and x = m, 1 mv2  q V …. (2) 2 v0  0 0 2q0 V  m  v0 = 3 m/s  Minimum value of v0 is 3 m/s. From equation (1) potential at origin (x = 0) is  8 1  V = 1.8 x 104     3 / 2  V = 2.4 x 104 V. Let K be the kinetic energy of the particle at origin. Applying energy conservation at x = 0 and at x =  . 1 K + q V = m v2 0 0 2 0 0 1 But from equation (2), m v2 = q V. 2 0 0 0  K = q0 (V – V0) K = 10-7 (2.7 x 104-2.4 x 104) K = 3 x 10-4 J. 3. A uniform electric field E is created between two parallel, charged plates as shown in figure. An electron enters the field symetrically between the plates with a speed u0. The length of each plate is 𝑙 . Find the angle of deviation of the path of the electron as it comes out of the field. eE V0 – – – – – – – – – – – – – – – – Solution: The acceleration of the electron is a = m in the upward direction. The horizontal velocity remains u0 as there is no acceleration in this direction. Thus, the time taken in crossing the field is : t  l u 0 The upward component of the velocity of the electron as it emerges from the field region is eEl uy = at = mu0 The horizontal component of the velocity remains ux = u0. The angle  made by the resultant velocity with the original direction is given by tan   uy  eEl u mu2 Thus, the electron deviates by an angle   tan1 uy  eEl . u mu2 4. Figure shows an electric dipole formed by two particles fixed at the ends of a light rod of length l. The mass of each particle is m and the chargers are –q and +q. The system is placed in such a way that the dipole axis is parallel to a uniform electric field E that exists in the region. The dipole is slightly rotated about its centre and released. Show that for small angular displacement, the motion is angular simple harmonic and find its time period. E -q +q E Solution: Suppose, the dipole axis makes an angle  with the electric field at an instant. The magni- tude of the torque on it is | → | → → | P  E |  q l Esin  This torque will be restoring & tend to rotate the elipole back towards the electric field. Also, for small angular displacement sin    so that   q l E It the moment of inertia of the body about OA is I, the angular acceleration becomes. where      I 2  qlE I qlE  I   2 T  2  2  Now, moment of inertia of the system about the axis of rotation is  l 2 I  2m     ml 2 2 So, T  2 . 5. Two conducting plates A and B are placed parallel to each other. A is given a charge Q1 charge Q 2 . Find the distribution of charges on the four surfaces. and B a Solution: Consider a Gaussian surface as shown in figure. Two faces of this closed surface lie com- pletely 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 the inner surface of B. E=0 Q1 A Q2 B E=0 (a) Q1 -q P A +q -q B Q2 +q (b) 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 each plate in A. Now since electric field at P E   20 the Due to the charge Q1  q  Q  q 2A (downward), Due to the charge Due to the charge 0  q  q 2A0  q  q 2A0 (upward), (downward), And due to the charge Q2  q  Q2  q (upward). 2A 0 The net electric field at P due to all the four charged surfaces is (in the downward direction) Q1  q  q  q  Q2  q . 2A0 2A0 A0 2A0 As the point P is inside the conductor, this field should be zero. Hence, Q1  q  Q 2  q  0 or, q  Q1  Q2 2 ...(i) (Q1+Q2)/2 A (Q1 – Q2)/2 Thus, And Q  q  Q1  Q 2 1 2 Q  q  Q1  Q 2 2 2 ...(ii) ...(iii) –(Q1 –Q2)/2 B (Q1+Q2)/2 Using these equations, the distribution shown in the figure can be redrawn as in figure. This result is a special case of the following result. When charge conducting plate are placed parallel to each other, the two outermost surfaces get equal charges and the facing surfaces get equal and opposite charges. 6. Figure shows three concentric thin spherical shells A, B and C of radii a, b and c respectively. The shells A and C are given charges q and – q respectively and the shell B is earthed. Find the charges appearing on the surfaces of B and C. q'-q -q C (a) C (b) Solution: As shown in the previous worked out example, the inner surface of B must have a charge – q from the Gauss’s law. Suppose, the outer a surface of B has a charge q . The inner surface of C must have a charge – q from the Gauss’s law. As the net charge on C must be – q, its outer surface should have a charge q  q . The charge distribution is shown in figure. Potential of shall B due to the charge q on the surface of A.  q 40 b due to the charge – q on the inner surface of B   q 40 b due to the charge q on the outer surface of B  q 40 b due to the charge  q , on the inner surface of C   q 40c and due to the charge q  q on the outer surface of C -bq/c  q1  b  q  q  c  bq  40c a q c -q The net potential is A VB  q  40 b q B 40c . C This should be zero as the shall B is earthed. Thus, q  b q . c The charges on various surfaces are as shown in figure 7. A cube of edge a metres carries a point charge q at each corner. Calculate the resultant force on any one of the charges. Solution: Let us take one corner of cube as origin O(0, 0, 0) and the opposite corner as P(a, a, a). We will calculate the electric field at P due to the other seven charges at corners. C P E A X F B O D Expressing the field of a point charge in vector form → q E  r 40 r 3 (i) Field at P due to A, B, C E  q AP  BP  CP 40a 3  q aˆj  akˆ  aˆi 40a 3 (ii) Field at P due to D, E, F Now that DP = EP = FP = a E  q DP  EP  FP 40 a 2  2 3  q aˆj  akˆ  aˆi  aˆj aˆi  akˆ  40 (2 2a 3 )  q ˆi  ˆj  kˆ 40 2a 2 (iii) Field at P due to O OP  a E3  E3  q 40 a 3 3 OP (aˆi  aˆj  akˆ) E  q ˆi  ˆj  kˆ  3 4 (3 3a 2 ) Resultant Field at P E  E1  E2  E3 q(ˆi  ˆj  kˆ)  E  40a 2 1   outward along OP Force on charge at P is F = q E q2 3   F  4 a2 1   Outwards along diagonal OP Note: In this problem, we have non-coplanar point charges and hence it is best to use vector approach in general form. 8. A uniform line charge  (in coulombs per meter) exists along the X-axis from x  a Find the electric field E at point P a distance r along the perpendicular bisector. to x  a . Solution: In all the problems which involve distributions of charge, we choose an element of charge dq to find the element of field dE’ produced at the given location. The we sum all such dE’s to find the total field E at that location. You must note the symmetry of the situation. For each element dq located at positive X, there is a similar dq(see mirror-image in origin) located at the same negative value of x. The dE x in the opposite direction due to the other dq. Hence, as we sum all the dq’s along the line, all the dE x components add to zero. So we need to sum only the dE y components, a scalar sum since they all point in the same direction. The element of charge is dq   dx . dE  1 40 dq r 2  x 2   dx 40 (r 2  x 2 ) Ex   dE x   dE sin   0 E y   dE y   dE cos  (by symmetry)  dx r  r  dx 40 (r 2  x 2 ) 40 (r 2  x 2 )3 / 2 The integral on the right hand side can be evaluated by substituting x = r tan  and dx  r sec 2  d  dx   r sec 2  d   cos  d  sin  (r 2  x 2 )3 / 2 r 3 sec 3  r 2 r 2  dx  1 x  (r 2  x 2 )3 / 2 r 2 E y  a  r 2a a The net field at P is E = E y E  Note: In case of an infinite lien charge, the field is everywhere perpendicular to the line of charge. The field at a distance r from the line is calculated by taking a   in the above result. E (infinite line charge)  lim  a a 20 r  E   20 r . 9. A system consists of a ball of radius R carrying a spherically symmetric charge and the surrounding space filled with a charge of volume density   a / r where a is a constant, r is the distance from the centre of ball. Find the ball’s charge at which the magnitude of the electric field is independent of r outside the ball . How high is this strength? Solution: Let us consider a spherical surface of radius Gauss’s Law. r(r  R) concentric with the ball and apply → → q □ E.dS  0 Let Q= total charge of the ball r 0 E(4r2 )  Q  4x2 dx R r a  E(4r )  Q  4 x 2 dx R x 0 E(4r 2 )  Q  2a(r 2  R 2 )  Q  2aR 2  1 2a  E    40   r 2  40 For E to be independent of r, Q  2aR 2 and the value of E is E  a 2 0 10. A charge Q is uniformly distributed over a spherical volume of radius R. Obtain an expression for the energy of the system. Solution: In this case, the electric field exists from centre of the sphere to infinity. Potential energy is stored in electric field with energy density. u  1 ε 2 0 E 2 (energy / volume) (i) Energy stored within the sphere (V1) Energy field at a distance r is (r  R) E  1 4πε0 . Q .r R 3 u  1 ε 2 0 E 2   1 Q 2 u  0  2  40 . R3 r Volume of element, dV = ( 4r2 )dr  Energy stored in this volume, dU = u(dV) dU = 4πr 2 dr ε 0  1  . Q r dU = 1 8πε 0 2 Q 2 . R 6 .r  4πε 0 R 3  4dr R 1 Q 2 R 4 U1 = 0 dU  8πε R 6 0 f dr  Q 2 40πε0 R 6 1 r 5 R 0 Q2 U1  40πε . R …. (1) (ii) Energy stored outside the sphere (U2) Electric field at a distance r is (r  R) E  1 . Q 4πε 0 r 2 u  1 ε 2 0 E 2 ε  1 Q 2 u  0  .  2  4πε 0 r 2  dV = (4  r2 dr) ε  1 dU = udV = (4  r2 dr)  0   2  4πε 0  2  .   r 2   dU  Q2 80  . dr r2 Q 2  dr U2 = R dV  8πε 0 r 2 U2 = Q2 8πε 0 R … (2) Therefore, total energy of the system is Q 2  Q 2 U = U1 + U2 = 40πε 0 R 8πε 0 R 3 Q2 U = 20 πε R . 11. Two particles of mass m and 2m carry a charge q each. Initially the heavier particle is at rest on a smooth horizontal plane and the other is projected along the plane directly towards the first from a distance d with speed u. Find the closest distance of approach. Solution: As the mass 2m is not fixed, it will also move away from m due to repulsion. The distance between the particles is minimum when their relative velocity is zero i.e., when they have equal velocities. d Hence at closest approach, By conservation of momentum mu  mv1  2mv 2 v 2  v1  u / 3 By conservation of energy Loss in KE = gain in PE v1  v2 at rest 1 2  1 2 1 2  q 2  1 1  2 mu   2 mv1  2 2mv 2   4    x d  1 2 1 u 2  0   q 2  1 1  mu  m 2 2 (1  2)  9    40  x d  1 mu 2  q  1 1     3 40  x d  1  1  40mu 2 x d 3q 2 3q2d x  3q2  4 mu2 d . A 12. The capacitance of a parallel plate capacitor with plate area A and separation d is C. The space between the plates is filled with two wedges of dielectric constants K1 and K2 respectively (Fig.). Find the capacitance of the resulting capacitor. Solution: Let length and breadth of the capacitor be l and b re- spectively and d be the distance between the plates as shown in fig. Then consider a strip at a distance x of width dx. Now QR = x tan  l and PQ = d – x tan  Where tan  = d/l, b Capacitance of PQ dC  k10 (b dx)  k10 (bdx) 1 d  x tan  d  xd l dC1  k10 bldx  k10 A(dx) P d(l  x) d(l  x) and dC2 = capacitance of QR dC2 dC2  k20b(dx) d tan   k20 A(dx) x d ......  tan   d  l   Now dC1 and dC2 are in series. Therefore, their resultant capacity dC will be given by 1  1  1 dC dC1 dC 2 then 1  1  1 dC dC1 dC 2  dl  x  x.d K1ε 0 Adx K 2 ε 0 Adx 1  d  l  x  x  d[K2 (l  x)  K1x] dC  A(dx)  K K  =  AK K (dx) 0  1 2  0 1 2 dC  0 AK1K2 dx d[K2 (l  x)  K1x] dC  0 AK1K2 dx d[K2l+(K1  K1)x] All such elemental capacitor representing DC are connected in parallel. Now the capacitance of the given parallel plate capacitor is obtained by adding such infinitesimal capacitors parallel from x = 0 to x = l. xl i.e. C   dC x0 l  AK K  0 1 2 dx 0 d[K2l  (K1  K2 )x] K1K 2 ε 0 A C = K1  K 2 d In K 2 . K1 13. The connections shown in figure are established with the switch S open. How much charge will flow through the switch if it is closed? (a) (b) Solution: When the switch is open, capacitors (2) and (3) are in series. Their equivalent capacitance is 2 F . The charge appearing on each of these capacitors is, therefore, 3 24V  2 F  16C . 3 The equivalent capacitance of (1) and (4), which are also connected in series, is also 2 F 3 and the charge on each of these capacitors is also 16C . The total charge on the two plates of (1) and (4) connected to the switch is, therefore, zero. The situation when the switch S is closed is shown in figure. Let the charges be distributed as shown in the figure. Q1 and Q 2 are arbitrarily chosen for the positive plates of (1) and (2). The same magnitude of charges will appear at the negative plates (3) and (4). Take the potential at the negative terminal to the zero and at the switch to be V0 . Writing equations for the capacitors (1), (2), (3) and (4). Q1  (24V  V0 ) 1 F Q2  (24V  V0 )  2F Q1  V0 1 F Q2  V0  2F From (i) and (iii), V0  12V . Thus, from (iii) and (iv), Q1  12C and Q2  24 C . ...(i) ...(ii) ...(iii) ...(iv) The charge on the two plates of (1) and (4) which are connected to the switch is, therefore Q2  Q1  12C . When the switch was open, this charge was zero. Thus, 12C passed through the switch after it was closed. of charge has 14. Each of the three plates shown in figure has an area of 200 cm2 on one side and the gap between the adjacent plates is 0.2 mm. The emf of the battery is 20 V. Find the distribution of charge on various surfaces of the plates. What is the equivalent capacitance of the system between the terminal points? (a) (b) Solution: Suppose the negative terminal of the battery gives a charge – Q to the plate B. As the situation is symmetric on the two sides of B, the two faces of the plate B will share equal charge –Q/ 2 each. From Gauss’s law, the facing surfaces will have charge Q/2 each. As the positive terminal of the battery has supplied just this much charge (+Q) to A and C, the outer surfaces of A and C will have no charge. The distribution will be as shown in figure. The capacitance between the plates A and B is C  A0  8.85 10 d 12 F / m  200 10 4 m2 2 104 m  8.85 10 10 F  0.885 nF . Thus, Q= 0885 nF × 20 V = 17.7 nC. The distribution of charge on various surfaces may be written from figure The equivalent capacitance is Q  1.77 nF 20 V 15. Find the capacitance of the infinite ladder shown in figure. P A B Q Solution: As the ladder is infinitely long, the capacitance of the ladder to the right of the points P, Q is the same as that of the ladder to the right of the points A, B. If the equivalent capacitance of the ladder is C1 , the given ladder may be replaced by the connections shown in figure. P A B Q The equivalent capacitance between A and B is easily found to be C  to the original ladder, the equivalent capacitance is also C1 . CC1 C  C1 . But being equivalent Thus, C1  C  CC1 C  C1 Or, C C  C2  C2  2CC Or, C2  CC1 C   C2  0 C 2  4C 2 1  5 C1 C2 Giving C1   C . 2 2 Negative value of C1 is rejected. 16. The emf of the cell in the circuit is 12 volts and the capacitors are : C1  1 μf , C2  3 μf , C3  2 μf , C4  4 μf. Calculate the charge on each capacitor and the total charge drawn from the cell when (a) the switch s is closed (b) the switch s is open. Solution:(a) Switch S is closed : C1  C3 C2  C4  C1  C3   C2  C4  C  3 7  2.1F 3  7 total charge drawn from the cell is : Q = C V = 2.1 μF × 12 volts = 25.2 μC C1, C3 are in parallel and C2, C4 are in parallel. Charge on C1 12 V C1 C2 Q1  C1 C1  C3 Q  1 1  2  25.2μC  8.4μC. Charge on C3 Q3  C3 C1  C3 Q  2 1  2  25.2μC  16.8μC. Charge on C2 Q2  C2 C2  C4 Q  3 3  4  25.2μC  10.8μC. Charge on C4 Q4  C4 C2  C4 Q  4 3  4  25.2μC  14.4μC. C1 C2 (b) Switch S is open : C  C1C2 C1  C2  C3C4 C3  C4 C  1 3  2  4  25 μF 1  3 2  4 12 total charge drawn from battery is : Q  CV  25 12  25μC 12 C1 & C2 are in series and the potential difference across combination is 12 volts. charge on C1 = charge on C2  C1C2  3    C1 V  12  9μC.  C2  4 C3 & C4 are in series and the potential difference across combination is 12 volts. charge on C3 = charge on C4  C3C4  8    C3 V  12  16μC.  C4  6 17. Two capacitors C1  1μF and C2  4μF are charged to a potential difference of 100 volts and 200 volts respectively. The charged capacitors are now connected to each other with terminals of oppo- site sign connected together. What is the (a) final charge on each capacitor in steady state ? (b) decrease in the energy of the system ? Solution: V1 V C C2 – – C – – C V2 Initial charge on C1  C1V1  100μC Initial charge on C2  C2 V2  800μC C1V1  C2 V2 when the terminals of opposite polarity are connected together, the magnitude of net charge finally is equal to the difference of magnitude of charges before connection. (charge on C2)i - (charge on C1)i = (charge on C2)f - (charge on C1)f Let V be the final common potential difference across each. The charges will be redistributed and the system attains a steady state when potential difference across each capacitor becomes same. C2 V2  C1V1  C2 V  C1V V  C1V2  C1V1  800 100  140 volts C2  C1 5 Note that because tion. C1V1  C2 V2 , the final charge polarities are same as that of C2 before connec- Final charge on C1 = C1 V = 140 μC Final charge on C2 = C2 V = 560 μC Loss of energy = Ui – Uf, Loss of energy = 1 C V2  1 C V 2  1 C V2  1 C V2 2 1 1 2 2 2 2 1 2 2  1 11002  1 42002  1 1  41402 2 2 2  36000μJ  0.036J Note: The energy is lost as heat in the connected wires due to the temporary currents that flow while the charge is being redistributed. 18. Four identical metal plates are located in air at equal separations d A as shown. The area of each plate is A. Calculate the effective capacitance of the arrangement across A and B. B Solution : Let us call the isolated plate as P. A capacitor is formed by a pair of parallel plates facing each other. Hence we have three capacitor formed by the pairs (1, 2), (3, 4) and (5, 6). The surface 2 and 3 are at same potential as that of A. The arrangement can be redrawn as a network of three capacitors. 1 A 2 1 3 4 5 6 P 5 6 B C  2C.C  2C AB 2C  C 3  2 ε 0A 3 d 19. Two capacitors A and B with capacities 3  F and 2F are charged to a potential difference of 100V and 180V respectively. The plates of the ca- pacitors are connected as shown in the figure with one wire of each capaci- + 2F - tor free. The upper plate of A is positive and that of B is negative. An uncharged 2  F capacitor C with lead wires falls on the free ends to com- plete the circuit. Calculate : (i) the final charge on the three capacitors, and 3F 2F A B (ii) the amount of electrostatic energy stored in the system before and afterthe completion of the circuit. Solution(i): Charge on capacitor A, before joining with an uncharged capacitor, q = CV = (100) × 3  c = 300  c similarly charge on capacitor B, Let q , q and q q =180 × 2  c = 360  c be the charges on the three capacitors after joining q2 B + - C 1 2 3 3 4 them as shown in fig. From conservation of charge, Net charge on plates 2 and 3 before joining = Net charge after joining + 2 3F- q 5 -2F + q3  300 = q + q2 …. (1) A D Similarly, net charge on plates 4 and 5 before joining = Net charge after joining -360 = - q2 – q3 360 = q2 + q3 …. (2) applying Kirchoff’s 2nd law in loop ABCDA, q1  q 2  q 3  0 3 2 2 2q1 - 3q2 + 3q3 = 0 …. (3) From equations (1), (2) and (3), q = 90  c, q = 90  c and q = 150  c (ii) (a) Electrostatic energy stored before completing the circuit, Ui = 1 2 (3 x 10-6 ) (100)2 1 + 2 (2  10-6 )(180)2 (U  1 CV 2 ) 2 = 4.74  10-2 J = 47.4 mJ. (b) Electrostatic energy stored after completing the circuit, U  1 (90 106 )2. 1  1 (90 106 )2. 1  1 (150  106 )2. 1 f 2 3106 2 2 106 2 2 106  U  1 q   2 c2    = 90 104 J = 9 mJ. 20. Two fixed positive charges, each of magnitude 5 105 C are lo- cated at points A and B, separated by a distance of 6 m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of lien AB. The moving charge, when reaches the point C at a distance of 4 m from O, has a kinetic energy of 4 joules. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C. Solution: The kinetic energy is lost and converted to electrostatic potential energy of the system as the negative charge goes from C to D and comes to rest at D instantaneously. Loss of KE = gain in PE 4  Uf  Ui  q q 2q(q)   qq 2q(q)  4          40 (6) 40 9  x2  40 (6) 40 9 16  2q 2 1 1  4  4 5   9  x 2  4  2(5 105 )2 (9 109 )  1  1    SOLVED PROBLEMS (SUBJECTIVE) 1. Three point charges q, 2q and 8q are to be placed on a 9 cm long straight line. Find the positions where the charges should be placed such that the potential energy of this system is minimum. In this situation, what is the electric field at the charge q due to the other two charges ? Solution: The maximum contribution may come from the charge 8q forming pairs with others. To reduce its effect, it should be placed at a corner and the smallest charge q in the middle. This arrangement shown in figure ensures that the charges in the strongest pair 2q, 8q are at the largest separation. The potential energy is 2q q 8q q2  2 16 8  U    . 40  x 9cm 9cm  x  This will be minimum if A  2  x 8 9cm  x is minimum. For this, dA   2  9cm  x2  0 …. (i) or, 9 cm – x = 2x or, x = 3 cm The electric field at the position of charge q is q  2 8      9cm  x2  = 0. From (i). 2. Four point charges +8  C, -1  C, -1  C and +8  C are fixed at the points - m, - m, + 3 / 2 m and + m respectively on the y-axis. A particle of mass 6 x 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. 1/(4   ) = 9 x 109 Nm2 / C2. Solution : in the figure q = 1  C = 10-6 C, and q = + 0.1  C = 10-7 C -4 y Q = 8  C = 8 × 10-6 C. B Let P be any point at a distance x from origin. Then A AP = CP = C D BP = DP = 27 / 2  x 2 Electric potential at point P will be: V  2KQ  2Kq , where BP AP K  1 4πε 0  9x109 Nm 2 / C 2   9  8  10 6 10 6  V  2  9.0  10     … (1)  27  x 2 3  x 2   2 2   Electric field at P is : dV   1  27 3 / 2  1  3   E    1.8 104  8     x2   2x 1    x2  2x dx E = 0 on x-axis where 8    1 2  2   2  2    27  3 / 2  x 2  3 / 2   x 2   2   2    27 43 / 2 3 / 2 2 1 3 / 2 2   x   2    x   2   27  x 2    3  x 2    4   2   2  x   5 m 2 The least value of kinetic energy of the particles of affinity should be enough to take the particles upto x   because At x   For x > and For x < E = 0 Fe = 0 Fe is repulsive (towards positive x-axis) 5 / 2 m ; Fe is attractive (towards negative x-axis) Now, from equation (1), potential at x = ,    8 1     V = 1.8 x 104  5 3 5     2  V = 2.7 x 104 volt. 2  2  Applying energy conservation at x =  and x = m, 1 mv2  q V …. (2) 2 v0  0 0 2q0 V  m  v0 = 3 m/s  Minimum value of v0 is 3 m/s. From equation (1) potential at origin (x = 0) is  8 1  V = 1.8 x 104     3 / 2  V = 2.4 x 104 V. Let K be the kinetic energy of the particle at origin. Applying energy conservation at x = 0 and at x =  . 1 K + q V = m v2 0 0 2 0 0 1 But from equation (2), m v2 = q V. 2 0 0 0  K = q0 (V – V0) K = 10-7 (2.7 x 104-2.4 x 104) K = 3 x 10-4 J. 3. A uniform electric field E is created between two parallel, charged plates as shown in figure. An electron enters the field symetrically between the plates with a speed u0. The length of each plate is 𝑙 . Find the angle of deviation of the path of the electron as it comes out of the field. eE V0 – – – – – – – – – – – – – – – – Solution: The acceleration of the electron is a = m in the upward direction. The horizontal velocity remains u0 as there is no acceleration in this direction. Thus, the time taken in crossing the field is : t  l u 0 The upward component of the velocity of the electron as it emerges from the field region is eEl uy = at = mu0 The horizontal component of the velocity remains ux = u0. The angle  made by the resultant velocity with the original direction is given by tan   uy  eEl u mu2 Thus, the electron deviates by an angle   tan1 uy  eEl . u mu2 4. Figure shows an electric dipole formed by two particles fixed at the ends of a light rod of length l. The mass of each particle is m and the chargers are –q and +q. The system is placed in such a way that the dipole axis is parallel to a uniform electric field E that exists in the region. The dipole is slightly rotated about its centre and released. Show that for small angular displacement, the motion is angular simple harmonic and find its time period. E -q +q E Solution: Suppose, the dipole axis makes an angle  with the electric field at an instant. The magni- tude of the torque on it is | → | → → | P  E |  q l Esin  This torque will be restoring & tend to rotate the elipole back towards the electric field. Also, for small angular displacement sin    so that   q l E It the moment of inertia of the body about OA is I, the angular acceleration becomes. where      I 2  qlE I qlE  I   2 T  2  2  Now, moment of inertia of the system about the axis of rotation is  l 2 I  2m     ml 2 2 So, T  2 . 5. Two conducting plates A and B are placed parallel to each other. A is given a charge Q1 charge Q 2 . Find the distribution of charges on the four surfaces. and B a Solution: Consider a Gaussian surface as shown in figure. Two faces of this closed surface lie com- pletely 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 the inner surface of B. E=0 Q1 A Q2 B E=0 (a) Q1 -q P A +q -q B Q2 +q (b) 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 each plate in A. Now since electric field at P E   20 the Due to the charge Q1  q  Q  q 2A (downward), Due to the charge Due to the charge 0  q  q 2A0  q  q 2A0 (upward), (downward), And due to the charge Q2  q  Q2  q (upward). 2A 0 The net electric field at P due to all the four charged surfaces is (in the downward direction) Q1  q  q  q  Q2  q . 2A0 2A0 A0 2A0 As the point P is inside the conductor, this field should be zero. Hence, Q1  q  Q 2  q  0 or, q  Q1  Q2 2 ...(i) (Q1+Q2)/2 A (Q1 – Q2)/2 Thus, And Q  q  Q1  Q 2 1 2 Q  q  Q1  Q 2 2 2 ...(ii) ...(iii) –(Q1 –Q2)/2 B (Q1+Q2)/2 Using these equations, the distribution shown in the figure can be redrawn as in figure. This result is a special case of the following result. When charge conducting plate are placed parallel to each other, the two outermost surfaces get equal charges and the facing surfaces get equal and opposite charges. 6. Figure shows three concentric thin spherical shells A, B and C of radii a, b and c respectively. The shells A and C are given charges q and – q respectively and the shell B is earthed. Find the charges appearing on the surfaces of B and C. q'-q -q C (a) C (b) Solution: As shown in the previous worked out example, the inner surface of B must have a charge – q from the Gauss’s law. Suppose, the outer a surface of B has a charge q . The inner surface of C must have a charge – q from the Gauss’s law. As the net charge on C must be – q, its outer surface should have a charge q  q . The charge distribution is shown in figure. Potential of shall B due to the charge q on the surface of A.  q 40 b due to the charge – q on the inner surface of B   q 40 b due to the charge q on the outer surface of B  q 40 b due to the charge  q , on the inner surface of C   q 40c and due to the charge q  q on the outer surface of C -bq/c  q1  b  q  q  c  bq  40c a q c -q The net potential is A VB  q  40 b q B 40c . C This should be zero as the shall B is earthed. Thus, q  b q . c The charges on various surfaces are as shown in figure 7. A cube of edge a metres carries a point charge q at each corner. Calculate the resultant force on any one of the charges. Solution: Let us take one corner of cube as origin O(0, 0, 0) and the opposite corner as P(a, a, a). We will calculate the electric field at P due to the other seven charges at corners. C P E A X F B O D Expressing the field of a point charge in vector form → q E  r 40 r 3 (i) Field at P due to A, B, C E  q AP  BP  CP 40a 3  q aˆj  akˆ  aˆi 40a 3 (ii) Field at P due to D, E, F Now that DP = EP = FP = a E  q DP  EP  FP 40 a 2  2 3  q aˆj  akˆ  aˆi  aˆj aˆi  akˆ  40 (2 2a 3 )  q ˆi  ˆj  kˆ 40 2a 2 (iii) Field at P due to O OP  a E3  E3  q 40 a 3 3 OP (aˆi  aˆj  akˆ) E  q ˆi  ˆj  kˆ  3 4 (3 3a 2 ) Resultant Field at P E  E1  E2  E3 q(ˆi  ˆj  kˆ)  E  40a 2 1   outward along OP Force on charge at P is F = q E q2 3   F  4 a2 1   Outwards along diagonal OP Note: In this problem, we have non-coplanar point charges and hence it is best to use vector approach in general form. 8. A uniform line charge  (in coulombs per meter) exists along the X-axis from x  a Find the electric field E at point P a distance r along the perpendicular bisector. to x  a . Solution: In all the problems which involve distributions of charge, we choose an element of charge dq to find the element of field dE’ produced at the given location. The we sum all such dE’s to find the total field E at that location. You must note the symmetry of the situation. For each element dq located at positive X, there is a similar dq(see mirror-image in origin) located at the same negative value of x. The dE x in the opposite direction due to the other dq. Hence, as we sum all the dq’s along the line, all the dE x components add to zero. So we need to sum only the dE y components, a scalar sum since they all point in the same direction. The element of charge is dq   dx . dE  1 40 dq r 2  x 2   dx 40 (r 2  x 2 ) Ex   dE x   dE sin   0 E y   dE y   dE cos  (by symmetry)  dx r  r  dx 40 (r 2  x 2 ) 40 (r 2  x 2 )3 / 2 The integral on the right hand side can be evaluated by substituting x = r tan  and dx  r sec 2  d  dx   r sec 2  d   cos  d  sin  (r 2  x 2 )3 / 2 r 3 sec 3  r 2 r 2  dx  1 x  (r 2  x 2 )3 / 2 r 2 E y  a  r 2a a The net field at P is E = E y E  Note: In case of an infinite lien charge, the field is everywhere perpendicular to the line of charge. The field at a distance r from the line is calculated by taking a   in the above result. E (infinite line charge)  lim  a a 20 r  E   20 r . 9. A system consists of a ball of radius R carrying a spherically symmetric charge and the surrounding space filled with a charge of volume density   a / r where a is a constant, r is the distance from the centre of ball. Find the ball’s charge at which the magnitude of the electric field is independent of r outside the ball . How high is this strength? Solution: Let us consider a spherical surface of radius Gauss’s Law. r(r  R) concentric with the ball and apply → → q □ E.dS  0 Let Q= total charge of the ball r 0 E(4r2 )  Q  4x2 dx R r a  E(4r )  Q  4 x 2 dx R x 0 E(4r 2 )  Q  2a(r 2  R 2 )  Q  2aR 2  1 2a  E    40   r 2  40 For E to be independent of r, Q  2aR 2 and the value of E is E  a 2 0 10. A charge Q is uniformly distributed over a spherical volume of radius R. Obtain an expression for the energy of the system. Solution: In this case, the electric field exists from centre of the sphere to infinity. Potential energy is stored in electric field with energy density. u  1 ε 2 0 E 2 (energy / volume) (i) Energy stored within the sphere (V1) Energy field at a distance r is (r  R) E  1 4πε0 . Q .r R 3 u  1 ε 2 0 E 2   1 Q 2 u  0  2  40 . R3 r Volume of element, dV = ( 4r2 )dr  Energy stored in this volume, dU = u(dV) dU = 4πr 2 dr ε 0  1  . Q r dU = 1 8πε 0 2 Q 2 . R 6 .r  4πε 0 R 3  4dr R 1 Q 2 R 4 U1 = 0 dU  8πε R 6 0 f dr  Q 2 40πε0 R 6 1 r 5 R 0 Q2 U1  40πε . R …. (1) (ii) Energy stored outside the sphere (U2) Electric field at a distance r is (r  R) E  1 . Q 4πε 0 r 2 u  1 ε 2 0 E 2 ε  1 Q 2 u  0  .  2  4πε 0 r 2  dV = (4  r2 dr) ε  1 dU = udV = (4  r2 dr)  0   2  4πε 0  2  .   r 2   dU  Q2 80  . dr r2 Q 2  dr U2 = R dV  8πε 0 r 2 U2 = Q2 8πε 0 R … (2) Therefore, total energy of the system is Q 2  Q 2 U = U1 + U2 = 40πε 0 R 8πε 0 R 3 Q2 U = 20 πε R . 11. Two particles of mass m and 2m carry a charge q each. Initially the heavier particle is at rest on a smooth horizontal plane and the other is projected along the plane directly towards the first from a distance d with speed u. Find the closest distance of approach. Solution: As the mass 2m is not fixed, it will also move away from m due to repulsion. The distance between the particles is minimum when their relative velocity is zero i.e., when they have equal velocities. d Hence at closest approach, By conservation of momentum mu  mv1  2mv 2 v 2  v1  u / 3 By conservation of energy Loss in KE = gain in PE v1  v2 at rest 1 2  1 2 1 2  q 2  1 1  2 mu   2 mv1  2 2mv 2   4    x d  1 2 1 u 2  0   q 2  1 1  mu  m 2 2 (1  2)  9    40  x d  1 mu 2  q  1 1     3 40  x d  1  1  40mu 2 x d 3q 2 3q2d x  3q2  4 mu2 d . A 12. The capacitance of a parallel plate capacitor with plate area A and separation d is C. The space between the plates is filled with two wedges of dielectric constants K1 and K2 respectively (Fig.). Find the capacitance of the resulting capacitor. Solution: Let length and breadth of the capacitor be l and b re- spectively and d be the distance between the plates as shown in fig. Then consider a strip at a distance x of width dx. Now QR = x tan  l and PQ = d – x tan  Where tan  = d/l, b Capacitance of PQ dC  k10 (b dx)  k10 (bdx) 1 d  x tan  d  xd l dC1  k10 bldx  k10 A(dx) P d(l  x) d(l  x) and dC2 = capacitance of QR dC2 dC2  k20b(dx) d tan   k20 A(dx) x d ......  tan   d  l   Now dC1 and dC2 are in series. Therefore, their resultant capacity dC will be given by 1  1  1 dC dC1 dC 2 then 1  1  1 dC dC1 dC 2  dl  x  x.d K1ε 0 Adx K 2 ε 0 Adx 1  d  l  x  x  d[K2 (l  x)  K1x] dC  A(dx)  K K  =  AK K (dx) 0  1 2  0 1 2 dC  0 AK1K2 dx d[K2 (l  x)  K1x] dC  0 AK1K2 dx d[K2l+(K1  K1)x] All such elemental capacitor representing DC are connected in parallel. Now the capacitance of the given parallel plate capacitor is obtained by adding such infinitesimal capacitors parallel from x = 0 to x = l. xl i.e. C   dC x0 l  AK K  0 1 2 dx 0 d[K2l  (K1  K2 )x] K1K 2 ε 0 A C = K1  K 2 d In K 2 . K1 13. The connections shown in figure are established with the switch S open. How much charge will flow through the switch if it is closed? (a) (b) Solution: When the switch is open, capacitors (2) and (3) are in series. Their equivalent capacitance is 2 F . The charge appearing on each of these capacitors is, therefore, 3 24V  2 F  16C . 3 The equivalent capacitance of (1) and (4), which are also connected in series, is also 2 F 3 and the charge on each of these capacitors is also 16C . The total charge on the two plates of (1) and (4) connected to the switch is, therefore, zero. The situation when the switch S is closed is shown in figure. Let the charges be distributed as shown in the figure. Q1 and Q 2 are arbitrarily chosen for the positive plates of (1) and (2). The same magnitude of charges will appear at the negative plates (3) and (4). Take the potential at the negative terminal to the zero and at the switch to be V0 . Writing equations for the capacitors (1), (2), (3) and (4). Q1  (24V  V0 ) 1 F Q2  (24V  V0 )  2F Q1  V0 1 F Q2  V0  2F From (i) and (iii), V0  12V . Thus, from (iii) and (iv), Q1  12C and Q2  24 C . ...(i) ...(ii) ...(iii) ...(iv) The charge on the two plates of (1) and (4) which are connected to the switch is, therefore Q2  Q1  12C . When the switch was open, this charge was zero. Thus, 12C passed through the switch after it was closed. of charge has 14. Each of the three plates shown in figure has an area of 200 cm2 on one side and the gap between the adjacent plates is 0.2 mm. The emf of the battery is 20 V. Find the distribution of charge on various surfaces of the plates. What is the equivalent capacitance of the system between the terminal points? (a) (b) Solution: Suppose the negative terminal of the battery gives a charge – Q to the plate B. As the situation is symmetric on the two sides of B, the two faces of the plate B will share equal charge –Q/ 2 each. From Gauss’s law, the facing surfaces will have charge Q/2 each. As the positive terminal of the battery has supplied just this much charge (+Q) to A and C, the outer surfaces of A and C will have no charge. The distribution will be as shown in figure. The capacitance between the plates A and B is C  A0  8.85 10 d 12 F / m  200 10 4 m2 2 104 m  8.85 10 10 F  0.885 nF . Thus, Q= 0885 nF × 20 V = 17.7 nC. The distribution of charge on various surfaces may be written from figure The equivalent capacitance is Q  1.77 nF 20 V 15. Find the capacitance of the infinite ladder shown in figure. P A B Q Solution: As the ladder is infinitely long, the capacitance of the ladder to the right of the points P, Q is the same as that of the ladder to the right of the points A, B. If the equivalent capacitance of the ladder is C1 , the given ladder may be replaced by the connections shown in figure. P A B Q The equivalent capacitance between A and B is easily found to be C  to the original ladder, the equivalent capacitance is also C1 . CC1 C  C1 . But being equivalent Thus, C1  C  CC1 C  C1 Or, C C  C2  C2  2CC Or, C2  CC1 C   C2  0 C 2  4C 2 1  5 C1 C2 Giving C1   C . 2 2 Negative value of C1 is rejected. 16. The emf of the cell in the circuit is 12 volts and the capacitors are : C1  1 μf , C2  3 μf , C3  2 μf , C4  4 μf. Calculate the charge on each capacitor and the total charge drawn from the cell when (a) the switch s is closed (b) the switch s is open. Solution:(a) Switch S is closed : C1  C3 C2  C4  C1  C3   C2  C4  C  3 7  2.1F 3  7 total charge drawn from the cell is : Q = C V = 2.1 μF × 12 volts = 25.2 μC C1, C3 are in parallel and C2, C4 are in parallel. Charge on C1 12 V C1 C2 Q1  C1 C1  C3 Q  1 1  2  25.2μC  8.4μC. Charge on C3 Q3  C3 C1  C3 Q  2 1  2  25.2μC  16.8μC. Charge on C2 Q2  C2 C2  C4 Q  3 3  4  25.2μC  10.8μC. Charge on C4 Q4  C4 C2  C4 Q  4 3  4  25.2μC  14.4μC. C1 C2 (b) Switch S is open : C  C1C2 C1  C2  C3C4 C3  C4 C  1 3  2  4  25 μF 1  3 2  4 12 total charge drawn from battery is : Q  CV  25 12  25μC 12 C1 & C2 are in series and the potential difference across combination is 12 volts. charge on C1 = charge on C2  C1C2  3    C1 V  12  9μC.  C2  4 C3 & C4 are in series and the potential difference across combination is 12 volts. charge on C3 = charge on C4  C3C4  8    C3 V  12  16μC.  C4  6 17. Two capacitors C1  1μF and C2  4μF are charged to a potential difference of 100 volts and 200 volts respectively. The charged capacitors are now connected to each other with terminals of oppo- site sign connected together. What is the (a) final charge on each capacitor in steady state ? (b) decrease in the energy of the system ? Solution: V1 V C C2 – – C – – C V2 Initial charge on C1  C1V1  100μC Initial charge on C2  C2 V2  800μC C1V1  C2 V2 when the terminals of opposite polarity are connected together, the magnitude of net charge finally is equal to the difference of magnitude of charges before connection. (charge on C2)i - (charge on C1)i = (charge on C2)f - (charge on C1)f Let V be the final common potential difference across each. The charges will be redistributed and the system attains a steady state when potential difference across each capacitor becomes same. C2 V2  C1V1  C2 V  C1V V  C1V2  C1V1  800 100  140 volts C2  C1 5 Note that because tion. C1V1  C2 V2 , the final charge polarities are same as that of C2 before connec- Final charge on C1 = C1 V = 140 μC Final charge on C2 = C2 V = 560 μC Loss of energy = Ui – Uf, Loss of energy = 1 C V2  1 C V 2  1 C V2  1 C V2 2 1 1 2 2 2 2 1 2 2  1 11002  1 42002  1 1  41402 2 2 2  36000μJ  0.036J Note: The energy is lost as heat in the connected wires due to the temporary currents that flow while the charge is being redistributed. 18. Four identical metal plates are located in air at equal separations d A as shown. The area of each plate is A. Calculate the effective capacitance of the arrangement across A and B. B Solution : Let us call the isolated plate as P. A capacitor is formed by a pair of parallel plates facing each other. Hence we have three capacitor formed by the pairs (1, 2), (3, 4) and (5, 6). The surface 2 and 3 are at same potential as that of A. The arrangement can be redrawn as a network of three capacitors. 1 A 2 1 3 4 5 6 P 5 6 B C  2C.C  2C AB 2C  C 3  2 ε 0A 3 d 19. Two capacitors A and B with capacities 3  F and 2F are charged to a potential difference of 100V and 180V respectively. The plates of the ca- pacitors are connected as shown in the figure with one wire of each capaci- + 2F - tor free. The upper plate of A is positive and that of B is negative. An uncharged 2  F capacitor C with lead wires falls on the free ends to com- plete the circuit. Calculate : (i) the final charge on the three capacitors, and 3F 2F A B (ii) the amount of electrostatic energy stored in the system before and afterthe completion of the circuit. Solution(i): Charge on capacitor A, before joining with an uncharged capacitor, q = CV = (100) × 3  c = 300  c similarly charge on capacitor B, Let q , q and q q =180 × 2  c = 360  c be the charges on the three capacitors after joining q2 B + - C 1 2 3 3 4 them as shown in fig. From conservation of charge, Net charge on plates 2 and 3 before joining = Net charge after joining + 2 3F- q 5 -2F + q3  300 = q + q2 …. (1) A D Similarly, net charge on plates 4 and 5 before joining = Net charge after joining -360 = - q2 – q3 360 = q2 + q3 …. (2) applying Kirchoff’s 2nd law in loop ABCDA, q1  q 2  q 3  0 3 2 2 2q1 - 3q2 + 3q3 = 0 …. (3) From equations (1), (2) and (3), q = 90  c, q = 90  c and q = 150  c (ii) (a) Electrostatic energy stored before completing the circuit, Ui = 1 2 (3 x 10-6 ) (100)2 1 + 2 (2  10-6 )(180)2 (U  1 CV 2 ) 2 = 4.74  10-2 J = 47.4 mJ. (b) Electrostatic energy stored after completing the circuit, U  1 (90 106 )2. 1  1 (90 106 )2. 1  1 (150  106 )2. 1 f 2 3106 2 2 106 2 2 106  U  1 q   2 c2    = 90 104 J = 9 mJ. 20. Two fixed positive charges, each of magnitude 5 105 C are lo- cated at points A and B, separated by a distance of 6 m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of lien AB. The moving charge, when reaches the point C at a distance of 4 m from O, has a kinetic energy of 4 joules. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C. Solution: The kinetic energy is lost and converted to electrostatic potential energy of the system as the negative charge goes from C to D and comes to rest at D instantaneously. Loss of KE = gain in PE 4  Uf  Ui  q q 2q(q)   qq 2q(q)  4          40 (6) 40 9  x2  40 (6) 40 9 16  2q 2 1 1  4  4 5   9  x 2  4  2(5 105 )2 (9 109 )  1  1     5  4  9  45 x   8.48 m .  5  4  9  45 x   8.48 m .