2.Electrotatics Solutions

2. ELECTROSTATICS 14.1 (D) R = = 8 N 8N 8N 14.2 (B) We can consider all the charge inside the sphere to be concentrated on the centre of sphere Consider an elementry shell of radius x and thickness dx. E = K  dq r 2 K 4 x2dx(ax) = r2 = r 2 4 0 14.3 (A) We have centripetal force equation q  2k  mv2  r  =   so v = 1 r Now, T = 2r v = where, k = 40 –→ → –→  ˆi q   (x  x ) 14.4 (A) Wnet = q E.d where E = 20 = 20 1 2 14.5 (A) Electric field between the two cylinders = 2kq 2k r \ Force on charge q = r This force is centrepetal force 2kq \ r = mv2 r \ v = = 14.6 (A) V = V1 + V2 + V3 = 1  Q  1  2Q   1  3Q  = 1   2Q   4 R 4  R  4  R  4  R  0 0   0   0   14.7 (B) dV = v → → E.dr = 1 (–2x3ˆi ) . (dx ˆi  dy ˆj  dzkˆ) = 2x3 dx Þ  dV = 0 (2x3 )  103 dx Þ V = – 7.5 × 103 V 2 14.8 (A) potential difference due to inner 10C charg  1 1  = K 10     = 9 × 1010 (5) = 45 × 1010 = 4.5 = 4.5 × 1011V 20C .1 .2 potential difference due to outer charge = 0 \ p.d. = 4.5 × 1011 V 14.9 (A) vA – vB = work done by electric field on + 1 coul. charge from A to B = E Rq \ vB = vA – E Rq = v – E Rq 14.10 (B) By argument of symmetry (it will be half of the potential produced by the full sphere) Þ 1  K(2Q) 2 R KQ KQ = R 9  109  5  109 V = R = 15  102 = 300 V 14.11 (B) Net potential of the sphere due to the induced charge is zero. Therefore, potential is due to the point charge only, equal to potential at the centre of the sphere. 14.12 (B) The given point is at axis of P 2 → dipole and at equatorial line of P dipole so total field at given point is. → =  kP  2k(P / 2) = →   1  7P = E (1)3 (2)3 kP  –1    32  0 k p E 14.13 (B) Electric field at each point on the surface of ring due to dipole is E = R 3 in direction opposite to the dipole moment. (figure below) kpQ Hence, net force on ring is F = QE = R3 Alternate solution Electric field due to ring at point P on its axis at a distance x from centre O of ring is E = k  Qx ; dE  at x  0 kQ Q = R3 \ Force on dipole = dE dx = k Qp R3 14.14 (C) After covering with a hemispherical shell; fshell + fshell = 0 (from Gauss law) \ fshell + fshell = – f 14.15 (B) eE = me w2– r m 2 R m 2R2 Edr = e  r dr Þ V = e e 0 2e 14.16 (B) U = KQ 1  E2 2 1 0K2Q2 2 2 r4 V = r 1  K2 Q2 U 2 0 r4 1 0 V2 = K2Q2 r2 = 2 2 r2 because U  1 V2 r2 so the correct option is B. 14.17 (C) f = → → → E  ds since r << R so we can consider electric field is constant throughtout the surface of smaller ring, hence f = E  x (R2  x2 )3 / 2 So, the best represented graph is C. 14.18 (C) Let q be positive. If it escapes then, from energy conservation principle, 1 mv2  KQq = 0 2 s R Vs v = {Note that Q is negative, therefore the quantity within the root is positive.} \ v  q When q is negative, escape velocity will be zero due to electrostatic repulsion from negative Q. → 14.19 (B) =    =  1 ˆi  1 ˆj  1 kˆ = ˆi  ˆj  kˆ E x2 y2 z2 14.20 (B) There exists a point P on the x-axis (other than the), where net electric field is zero. Once the charge Q reaches point P, attractive forces of the two –ve charge will dominate and automatically cause the charge Q to cross the origin. Now, if Q is projected with just enough velocity to reach P, its K.E. at P is zero. But while being attrached towards origin it acquires K.E and hence its net energy at the origin is positive. (P.E. at origin = zero). 14.21 (C) The electric field inside the inner shell is zero. So, the potential on inner shell and all the points inside it will be constant. 14.22 (B) V = Kp.r r3 9  109  (2ˆi  3ˆj  4kˆ). (2ˆi  2ˆj – kˆ) = [22  22  12 ]3 / 2 9  109  (4  6 – 4) = 27 V = 2 × 109 volts Ans. 14.23 (D) Interaction energy of system of charges is = 1 [U  ...  U ] = 1 [6U ] = 3U 2  2 1  q2 2 1 6 2 1 1  3  15  = 3 kq   a   2a  =   a  2 8    0   14.24 (A) The distribution of charge on the outer surface, depends only on the charges outside, and it distributes itself such that the net, electric field inside the outer surface due to the charge on outer surface and all the outer charges is zero. Similarly the distribution of charge on the inner surface, depends only on the charges inside the inner surface, and it distributes itself such that the net, electric field outside the inner surface due to the charge on inner surface and all the inner charges is zero. Also the force on charge inside the cavity is due to the charge on the inner surface. Hence, answer is option (A). 14.25 (C) Using the formula for electric field produced by large sheet E = We get; Q 2A 0 EA = 4Q 2A 0  (ˆi ) ; EB = 2Q 2A 0  (ˆi ) EC = 4Q 2A 0  (ˆi ) 14.26 (D) VB  VA   Exdx = – [Area under Ex – x curve] VB – 10 =  1 .2.(20) 2 = 20 Þ VB = 30 V. 14.27 (C) F = kq2 r2 Þ K = 1 4 0 Þ kq2 r2 mv2 = RC Þ Rc = mv 2r 2 kq2 RC = 4 0v2r2m q2 14.28 (B) Speed will be maximum when acceleration becomes zero. EQ i.e., when kx = EQ Þ X = K By work-energy theorum: W all = D KE Þ EQX  1 KX2 2 = 0 ; Xmax 2EQ = K 14.29 (B) OEH is an equipotential surface, the uniform E.F. must be perpendicular to it pointing from heigher to lower potential as shown Hence, Eˆ  ˆi  ˆj  =   (VE  VB ) Y D V=1 C H V=0 F V=–1 E = → \ E = EB = E. Eˆ = = = ˆi  ˆj . O A B X E 14.30 (C) Vp = ka  V r in kq = VC = r Þ Vin = kq  kq r r 14.31 (C) The electric field intensity due to each uniformly charged infinite plane is uniform. The electric field intensity at points A, B, C and D due to plane 1, plane 2 and both planes are given by E1, E2 and E as shown in figure 1. Hence, the electric lines of forces are as given in figure 2. E E B E2 1 C E2 E1 E z– E E 1 E2 A x E  D E E1 2 X (figure 2) (figure 1) Aliter: Electric lines of forces originate from positively charged plane and terminate at negatively charged plane. Hence, the correct representation of ELOF is as shown figure 2. 14.32 (C) The acceleration of centre of mass of system of particles is acm = (q1  q2 ) E 2m \ x-coordinate of centre of mass at t = 2 second is X = 1 a t2 = 1 q1  q2 E  22 = q1  q2 E cm 2 cm 2 2 m m Let the x-coordinates of q1 and q2 at t = 2 sec be x1 and x2 [x1 = 2a at t = 2 sec.] \ Xcm = mx1  mx2 2m = x1  x2 2 or x2 = 2 xcm – x1 = 2 (q1  q2 ) E  2a . m 14.33 (A) The bowl, exerts a normal force N on each bead, directed along the radius line or at 60° above the horizontal. Consider the free-body diagram of the bead on the left with the electric force Fe applied. S Fy = N sin 60° – mg = 0, Þ N = mg/sin 60° S Fx = – Fe + N cos 60° = 0, Þ 1 Kq2 R2 = N cos 60° = mg tan 60 = k = 4  x Thus q =  mg 1/ 2 R     14.34 (A) The potential at centre of sphere in which q charge is uniformly distributed throughout the volume is– 1 3q VC = 4 0 2R By symmetry the potential at centre due to half sphere will be half of the complete sphere. 1 3q/ 2 1 3Q  q  \ VC = 4 0 2R = 4 0 2R  2  Q 14.35 (A, B, D) Increasing the accelerating voltage means increasing speed of the electron, thereby decreasing time spend between the plates. It will reduce X. Increasing deflecting voltage means increasing electric field between the plates, making acceleration of electron greater. Increasing distance once again will change electric field between the plates. 14.36 (A, B, C) time of flight (t) = R 2u 2  10 g = 10 = 2 sec u2 10  10 1  qE  2 1 103  104  2  2 H = 2g = 2  10 = 5m Þ R = 0    t =    2 2 = 10 m 14.37 (A, B, C, D) Charge on a1 = (r1 q) l Charge on a2 = (r2 q) l r1 Ratio of charges = 2 E , Field produced by a = K [(r1 ) ] K Q  = 1 1 2 r1 K Q  E2, Field produced by a2 = r2 as r2 > r1 Therefore E1 > E2 i.e., Net field at A is towards a2. K.(r1) K.(r2) V1 = r1 = Kql Þ V2 = r2 = Kql Þ V1 = V2.  x  14.38 (A, B, D) 0 ³ x £ a : Vx =  Exdx   V(0) = 0 (as Ex = 0)  0  x  x    x ³ a £ Vx = Exdx  V(a) =    dx  V(a) =  (x  a) z  a 0  0 x      .x   .x x ³ a £ Vx = Exdx  V(0) = 0    0  V(0) = . 0 14.39 (B, C, D) V at origin ¹ 0 E(r = 2m) = K(q)r (R2  r2 )3 / 2  K.Q.r (R2  r2 )3 / 2 1 2  = K.rq   1 103 / 2   23 / 2.103 / 2  = 0 From origin to r = 2, field is towards origin. 14.40 (A, B, C) Charge is distributed over the surface of conductor in such a way that net field due to this charge and outside charge q is zero inside. Field due to only q is non-zero. 14.41 (D) Potential at a point is zero does not imply that electric field at same point should be zero. For instance in the equatorial plane of a dipole, potential at any point is zero but electric field is not zero. Hence, statement 1 is false. No electric field in space means, potential at all points in space is same. Hence, potential difference between any two points is zero. Hence, statement 2 is true. Statement-1 is False, Statement-2 is True. 14.42 (A) The electric field inside the cavity depends only on point charge q. Hence, VA – VB remains constant even if point charge Q is shifted. Here statement 2 is correct explanation of statement 1. 14.43 (D) For a non-uniformly charged thin circular ring with net zero charge, electric potential at each point on its axis is zero. Hence, electric field at each point on its axis must be perpendicular to the axis. Therefore, statement 1 is false and statement 2 is true. 14.44 (A) The electric field due to disc is superposition of electric field due to its constituent ring as given in statement-2. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. 14.45 (C) From work energy theorem Final K.E. = Initial K.E. + work done by non-uniform electric field. \ initial K.E. = 0 and final K.E. cannot be negative. \ Work done by non-uniform electric field on a charged particle starting from rest is non-negative. Hence, statement I is true. Consider a situation in which two point charges + Q are fixed some distance apart. At some distance left of equilibrium point O, a charge + q is released from rest. After the charge +q moving towards right crosses O, it experiences a force towards left. Hence, statement II is False. + Q +q O fixed charge Statement-1 is True, Statement-2 is False. + Q fixed charge Sol. (14.46 to 14.48) (B) (C) (A) The inner sphere is grounded, hence its potential is zero. The net charge on isolated outer sphere is zero. Let the charge on inner sphere be q. \ Potential at centre of inner sphere is The region in between conducting sphere and shell is shielded from charges on and outside the outer surface of shell. Hence, charge distribution on surface of sphere and inner surface of shell is uniform. The distribution of induced charge on outer surface of shell depends only on point charge q, hence is nonuniform. The charge distribution on all surfaces, is as shown. The charge distribution on all surfaces, as as shown. The electric field at B = 1 40  q 4X2 towards left. \ V = V a – V =   1 q dx = 1  q 32  a C C A 2a 40 4X2 0 14.49 (D) Let the speed of charges A and B be VA and VB when the separation between them is l0. Then from conservation of momentum. – mVA + 2 VB = 0 or VA = 2 VB Applying conservation of energy, as the separation increases from l0 to 2l0. Gain in K.E. of system of charges = Loss in electrostatic potential energy of the system of charges. 1 mV2  1 2mV2 1 2q2  1 2q2 2 A 2 B = 4 l 4 2l 1 2 1  V 2 0 0 0 0 1 q2 or mVA  2m  A  = 2 2  2  40 l0 Solving we get the speed of charge A is VA = . 14.50 (B) The work done by electrostatic force on charge A, from work energy theorem, in the given duration is = Final kinetic energy of charge A – Initial kinetic energy of charge A = 1 mVA  0 2 2 1 q2 = 6 . 0 l0 The sign of work done is positive. 14.51 (C) The net-work done by electrostatic force on system of two charged particle is equal to change in electrostatic potential energy of the system. 1 2q2 1 2q2 1 q2 = 4  l 4 2l = 4 l . The sign of work done is positive. 0 0 0 0 0 0 Sol. (14.52 to 14.54) (B) (C) (D) Potentials at the centre q/3 q/3 q/4 q/4 V = 1 q; V = 1 q 1 40 r 2 40 r Potential energy in situation I is 3 R 30° q/4 q/4 U1 = 3  = q/3 situation A situation B When one charge is removed, the field intensity at the centre is due to the removed charge only. 1 q/ 3 E1 = 40 r2 E2 = 1 q / 4 40 r2 E1 4 \ E2 = 3 14.55 (B) Consider two small elements of ring having charges + dq and – dx symmetrically located about y-axis. The potential due to this pair at any point on y-axis is zero. The sum of potential due to all such possible pairs is zero at all points on y-axis. Hence, potential at P 0, R  is zero.  2    y' 14.56 (A) Since, all charge lies in x-y plane, hence, direction of electric field at point P should be in x- y plane. Also y-axis is an equipotential (zero potential) line. Hence, direction of electric field at all points on y-axis should be normal to y-axis. \ The direction of electric field at P should be in x-y plane and normal to y-axis. Hence, direction of electric field is along positive-x-direction. 14.57 (C) Consider two small elements of ring having charge +dq and –dq as shown in figure. The pair constitutes a dipole of dipole moment. dq = dq 2R = (l Rd q) 2R The net dipole moment of system is vector sum of dipole moments of all such pairs of elementary charges. By symmetry the resultant dipole moment is along negative x-direction. \ net dipole moment = =  4R2 ˆi .  / 2    / 2 (dp cos ) ˆi =  / 2    / 2 (2R2 cos  d ) ˆi y' 1 (Q)2 Q2 14.58 (A) Electrostatic potential energy = = 40 2a 8 0 a 1 (Q)  (Q) (Q)2  3 Q2 (B) Electrostatic potential energy =  4 0  5a / 2  2(5a / 2)  = 20 0 a 1 3Q2 3 Q2 (C) Electrostatic potential energy = = 4 0 5a 20  0a 1  3Q2 (Q)2 (Q)  (Q)  27 Q2 (D) Electrostatic potential energy = Ans. (A) p, s (B) q, s (C) q, s (D) s 4 0  5a   2(2a)  = 2a  80  0 a 14.59 In situation A, B, and C, shells I and II are not at same potential. Hence, charge shall flow from sphere I to sphere II till both acquire same potential. If charge flows, the potential energy of system decreases and heat is produced. In situations A and B charges shall divide in some fixed ratio, but in situation C complete charge shall be tansferred to shell II for potential of shell I and II to be same. (A) ® P, q , (B) ® P, q , (C) ® P, q, s In situation D both the shells are at same potential, hence no charge flows through connecting wire. \ (D) ® r, s. Ans. (A) p, q (B) p, q (C) p, q, s (D) r, s, t 14.60 The path of the particle will be as shown in the figure. At the point of minimum distance (D) the velocity of the particle will be ^ to its position vector w.r. to +Q. Now by conservation of energy. d 1 mu2  0 = 2 1 mv2  KQq 2 rmin +Q ...(1) \ Torque on q about Q is zero hence angular momentum about Q will be conserved Þ m v rmin = m ud (2) by (2) in (1) Þ 1 mu2 = 2 1  ud 2 m   2  min   KQq rmin 1 mu 2  2 1   2   r 2  mu2d = rmin { \ KQq = mu2d (given)}  min  Þ r2  2r d  d2 = 0 Þ r = = d (1  2) min min \ distance cannot be negative min 2 \ rmin = d (1 ± 14.61 Flux through ABCD → → 2) Ans. Ans. 1 f1 = EA =(x2ˆi  yˆj) . (a2ˆi ) = 0 as x = 0 y Flux through EFGH f2 = (x ˆi  yˆj) . ( a ˆi) 2 2 = x2,a2 = a4 = 1.0 × 10–4 Nm2/C X Flux through BCGF f3 = (x ˆi  yˆj) . (a ˆj) 2 2 = a3 = 1,0 × 10–3 Nm2/C Flux through EADH f4 = (x2ˆi  yˆj) . (a2ˆj) = 0 as y = 0 Flux through ABFE f5 = (x ˆi  yˆj) . (a kˆ) = 0 2 2 Flux through CDHG f6 = 0 Net flux = (1.0 × 10–4 + 1.0 × 10–3) N-m2/C = 11 × 10–4 N-m2/C Ans. 11 14.62 At equilibrium mg = qE q2k mg = h ...(1)  when particle is displaced by a small distance x in upward direction then  2k  2kq  x 1 2kq  x  Net force F = q h  x   mg = 1      mg = 1    m   F =  2kq  x = – K x  h  0   \ F µ – x Motion is SHM Time period of motion 2 T = Þ T = 2 Ans. 14.63 \ Electric field near point b is – ¥ \ 'b' should be negative electric field at x1 is O which possible only if 'a' and 'b' are of opposite sign. \ 'a' is positive Charge b is negative and charge a is positive x1 A Qa Qb | Q | | Q | Q  l  x 2  l 2 a b E at A = 0 Þ = \ a =  1  = 1   (l  x1)2 (x1)2 Qb  x1   x1  E at a general X Þ K | Qa |  K | Qb | = K | Q | 1  1  (l  x2 ) (x2 ) dE 2  x a (l  x)2 2 2 x2  If E is a maximum, dx 2 = 0 Þ (l x)3   1   l  x1  2 x3 = 0 (l + x)3 = x3  l  x1  ; l + x = x  l  x1 3 l \ x =  x1   x1  2 2  l  x1 3   x1  Ans: x = l , Qa = 1  l 2  , Charge b is negative and charge a is positive 2 2 Qb  l  x1 3  1  x1     1  14.64 \ net effect due to charge inside the cavity is zero for outside point. On S1 surface -q charge distributed non-unifrmely and on S2 surface (Q + q) charge distributed unidormely. K.(Q  q) Ep = Vp = r2 K.(Q  q) r +q V = Kq  K(q)  K(Q  q) c x a b Ans. Ep = K(Q  q); r2 V = K(Q  q); p r V = Kq  K(q)  K(Q  q) c x a b 14.65 Assume a solid sphere without cavity.  4 R2 2 3 1  3  R potential at A due to this solid spehre = VA = .   = 2 40 R 2 0 Electric field 'C' due to this = → EC =  3 0 –––→ AC now consider the cavity filled with negative charge 4  R 3 A 1 3   2  () R2 VA = 4 0   R 2 = 12 0 → EC  = 3 0 –––→ BC Now net values for the solid sphere with the cavity can be given by superposition of the above two cases R2  1  1   5 R2 Hence, VA = VA  VA = 0  2 12  = 12 0 –→ –→ →  –––→ –––→  –––→ EC = E  E  = (AC  BC) = AB C C 3 0 3 0   R  R \ Ec = 3 0   = 6 0 Ans. VA = 5R2 , 12 0 EC = R 6 0