640 Electricity and Magnetism 109 ) q1 + q2 2 8. If we make a bigger cube comprising of eight (0.5)2 2 (9 × small cubes of size given in the question with charge at centre (or at D). 0.036 = …(ii) Solving Eqs. (i) and (ii), we can find q1 and q2. Then, total flux through large closed cube = q . ε0 6. Since, net force on electric dipole in uniform There are 24 symmetrical faces like EFGH on electric field is zero. Hence, torque can be outermost surface of this bigger cube. calculated about point. This comes out to be a constant quantity given by q Total flux from these 24 faces is . Hence, flux τ = p×E ε0 7. E2 from anyone force = 1 q . 24 ε0 E1 θ θ E1 Net = E P P Electric lines are tangential to face AEHD. Hence, flux is zero. r θθ r y 9. (a) –a +a 2 +a q a –2q a q q1 O Q q2 E1 = Electric field at P due to q q1 and q2 should be of same sign. = 1 q Further, Kq1Q = Kq2Q πε0 r2 (3a/ 2)2 (a/ 2)2 4 E2 = Electric field at P due to − 2q ∴ q1 = 9q2 = 1 2q –a +a + 3a 4 πε 0 y2 2 (b) q1 O q2 Q ∴ Net electric field at P, Therefore, q1 and q2 should be of opposite signs. Further, E = 2E1 cos θ + E2 = 1 rq2 ry + 1 2yq2 kq1Q = kq2Q 2 4πε0 4πε0 (5a/ 2)2 (a /2)2 = 2q y + 1 Or magnitude wise q1 = 25 q2 with sign 4 πε 0 r3 q1 = − 25q2 y2 xy = 2q y + 1 10. +q –Q +4q 4 πε 0 y2)3/ 2 (a2 + y2 x+ y=L …(i) …(ii) 2q y a2 − 3/ 2 1 Let force on − Q charge should be zero. 4 πε 0 y3 1 + y2 y2 = + + kQ. q kQ (4q) x2 y2 ∴ = Applying binomial expression for a2 < < 1 ∴ x=1 y2 y2 we get, E = 2q 3 a3 From Eqs. (i) and (ii), we get 4 πε 0 − 2 y4 x = L and y = 2L 33 = − 3qa2 4πε0 y4 Net force on 4q should be zero. or E = − 3qa2 $j ∴ k |Q | (4q) = kq (4q) 4πε0 y4 (2L / 3)2 L2
Chapter 24 Electrostatics 641 ∴ |Q | = 4 q Now, applying Lami’s theorem for the equilibrium With sign, 9 of ball we have Q =− 4q mg = 3F 9 sin (90° + 30° ) sin (90° + 60°) Similarly net force on q should be zero. F = mg = 1 (q) (q) = mg 3 4 πε 0 r2 3 ∴ k |Q| q = kq (4q) or |Q | = 4 q ∴ q = mgr2 (L / 3)2 L2 9 3 (1/4 πε0) If we, sightly displace − Q towards 4q, attraction = (0.1 × 10− 3) (9.8) (10 3 × 10− 2)2 between these charges will increase, hence − Q 3 × 9 × 109 will move towards + 4q and it will not return back. Hence, equilibrium is unstable. = 3.3 × 10− 8 C N 30° qaq 11. 60° F 13. a mg Using Lami’s theorem, we have q p E1 mg = F E2 E1 sin (90° + 30° ) sin (90° + 60°) ∴ F = mg or 1 (q) (q) = mg E1 = Electric field due to charge q at distance a 3 4 πε 0 R2 3 1 q ∴ q = 4π ε0 mg R2 = 4πε0 a2 3 E2 = Electric field due to charge at distance 2a 30° T = 1 ( q = E1 30° 4 πε 0 2 a)2 2 12. √3F Net electric field at P, mg Enet = 2 E1 + E2 (In the direction of E2) 2 F = Electrostatic force between two charged balls. 3F = Resultant of electrostatic force on any one = 2E1 + E1 2 ball from rest two balls. = 2 + 21 E1 30° = (2 2 + 1)q l 8πε0a2 60° 14. E= 1 q (rp − rq ) a 30° 4 πε 0 r3 r r = (1.2)2 + (1.6)2 = 2m a = l cos 60° = 10 cm (9 × 109 ) (− 8 × 10− 9) (2)3 r = 2 a cos 30° ∴ E = (1.2 i$ − 10.8i$) 3 = 10 3 cm = (14.4$j − 10.8$j) N/ C = (2) (10) 2
642 Electricity and Magnetism y dq dq 15. dθ +dq θ dE x θ θ Enet θθ (b) θ θθ O dE +dq dE dE –dq dE Enet Enet dE (c) 90° ∫Enet = 2 0 dE sin θ ∫= 2 90° 1 Rdq2 sin θ dq 0 4πε0 ∫= 2 90° 1 λRdθ sin θ = λ 18. Six vectors of equal magnitudes are as shown in 0 4π ε0 R2 2πε0R figure. Enet –12q –q dE θ θ dE –11q 16. θθ r –10q 1 –2q –9q 2 –3q a dx 3 –8q –4q dq dq 4 x 56 x = L/ 2 –7q –5q –6q ∫Enet = 2 x = 0 dE cos θs ∫= 2 L/ 2 1 dr2q ⋅ ar Now, resultant of two vectors of equal magnitudes 0 4πε0 (= E say ) at 120° is also E and passing through their bisector line. ∫=2 L/ 2 1 Q ⋅ dx a 0 4πε0 + x2 a2 + x2 L So, resultant of 1 and 5 is also E in the direction of a2 3. Similarly, resultant of 2 and 6 is also E in the direction of 4. Solving this integration, we find the result. Finally, resultant of 2E in the direction of 3 and 2E y in the direction of 4 passes through the bisector line of 3 and 4 (or 9.30) Enet −dq dE 19. T = 2u sin θ = 2 × 25 × sin 45° = 5 s 17. (a) θ x g 10 2 θ 1 2 R = Sx = uxT + 2 axT dq = (u cos 45°) (T ) + 1 qmE T2 2
Chapter 24 Electrostatics 643 = (25) 12 52 2 × 10−6 × 2 × 107 5 2 1 2 + 23. C = 62.5 + 250 = 312.5 m dq ra 2√3 20. (a) ge = g + qE m dx x (a/2) = 10 + (1.6 × 10− 19 ) (720) 1.67 × 10− 27 dq = λdx ≈ 6.9 × 1010 m/s2 Potential at C, due to charge dq is Now, R = u2 sin 2θ dV = k (dq) k = 1 ge r Q 4 πε 0 ∴ (1.27 × 10− 3) = (9.55 × 103)2 sin 2θ ∫∴ Total potential = 6 x = a/ 2 dV 6.9 × 1010 x=0 Solving this equation, we get 24. (a) ∆U = − q (∆V ) θ = 37° and 53° = − (12 × 10− 6) (50) (b) Apply T = 2u sin θ = − 6.0 × 10− 4J ge (b) ∆V = Ed = E (∆x) qE (1.6 × 10− 19 ) (120$j) 21. (a) a = m = − (9.1 × 10− 31) = (250) (0.2) = 50 V = − (2.1 × 1013$j) m/s2 25. kA + U A = kB + U B X 2 × 10− 2 4 (b) t = vx = 1.5 × 105 = 3 × 10− 7 s ∴ 1 mvA2 + qVA = 1 mvB2 + qVB 2 2 v = u + at ∴ vB = vA2 + 2 q (VA − VB ) m = (1.5 × 105 i$ + 3.0 × 106 $j) − 2.1 × 1013 × 4 × 10− 7 $j = (5)2 + 2 × (− 5 × 10− 6) (200 − 800) 3 2 × 10− 4 = (1.5i$ + 2$j) × 105 m/s = 7.42 m/s 22. kq1 = k |q2| vB − vA as the negative charge is moving (freely) from lower potential at A to higher potential at B. r1 r2 So, its electrostatic PE will decrease and kinetic ∴ r1 = q1 = 2 energy will increase. r2 |q2 | 3 where, r1 = distance from q1 and 26. (a) VC = k. qnet = 1 − 5Q r2 = distance from q2 R 4 πε 0 R 40 cm 60 cm (b) Vp = k qnet r q1 P q2 x = 0 x = 40 cm where, r = distance of P from any point on First point is at x = 40 cm where, r1 = 2 circumference r2 3 = 1 − 5Q 4πε0 R2 + z2 200 cm 100 cm P q1 q2 27. W = − ∆U = U i − U f = q2Vi − q2V f x = –200 cm x = 0 Second point is at x = − 200 cm, where r1 = 2. = q2 kq1 − q2 kq1 r2 3 ri rf
644 Electricity and Magnetism = kq1 q2 1 − 1 (c) Substituting, A = 10, x = 1, y = 1and z = 1in the ri rf expression of part (b) we have E = (− 20$i − 20$j − 20k$ ) N/C = (9 × 109 ) (2.4 × 10− 6) (− 4.3 × 106) ∴ |E | = (− 20)2 + (− 20)2 + (− 20)2 1 − 1 2 = 20 3 N/C 0.15 0.25 = − 0.356 J 34. See the hints of Q No. 31 to Q No. 33. q1q2 q1 q3 q2q3 V f − Vi = VB − VA = − 20 (xf − xi ) − 30 ( yf − yi ) r12 r13 r23 28. (a) U = k + + …(i) or Vf = − 20 (2 − 0) − 30 (2 − 0) where, k = 1 = − 100 V (as Vi = 0) 4 πε 0 35. (a) E = − ∂V i$ + ∂V $j + ∂V k$ ∂x ∂y ∂z (b) Suppose q3 is placed at coordinate x (> 0.2 m = − [(Ay − 2Bx) i$ + (Ax + C )$j] or 20 cm), then in Eq. (i) of part (a), put U = 0, r12 = 0.2 m, r13 = x ∴ Ex = (− Ay + 2Bx) and r23 = (x − 0.2) and Ey = (− Ax − C ) Now, solving Eq. (i) we get the desired value of x. (b) E = 0 if Ex and Ey are separately zero. 29. U = 0 Q − Ay + 2Bx = 0 …(i) ∴ k q × q + q×Q + q × Q =0 and − Ax − C = 0 …(ii) a a a Solving these two equations or Q = − q We get 2 x=−C and y = − 2BC A A2 ∫30. B Apply VB − VA = − …(i) 36. Sphere is a closed surface. Therefore, Gauss’s E ⋅ dr A E is given in the question. theorem can be applied directly on this and dr = dx$i + dy$j ∴ φ total = qin ε0 ∴ E⋅ dr = (5dx − 3dy) ∴ − ∫ E⋅ dr = (3 y − 5x) or qin = (φtotal ) (ε0) = (360) (8.86 × 10− 12) With limits answer comes out to be VB − VA = 3 ( yf − yi ) − 5 (xf − xi ) = 3.19 × 10−9 C 31. Procedure is same as work done in the above = 3.19 nC question. The only difference is, electric field is 37. (a) φ total = qin ε0 E = (400$j) V/m VB − VA = − B − − (b) qin = (φtotal ) (ε0) ∫∴ E⋅ dr = 400 ( yf yi ) A 38. S = (0.2) i$ 32. Similar to above two problems. But electric field ∴ φ = B ⋅S = 0.2 3 E0 5 here is E = (20$j) N/C B = 0.2 × 0.6 × 2 × 103 ∫∴ VB − VA = − A E ⋅ dr = − 20 (xf − xi ) = 240 N-m2/C 33. (a) [ A ] = [V ] = [ML2 T−3A−1 ] 39. Electric flux enters from the surface parallel to [xy] [L ][L ] y - z plane at x = 0. But E = 0 at x = 0. = [MT−3 A−1 ] Hence, flux entering the cube = 0. (b) E = − ∂V $i + ∂V $j + ∂V k$ Flux leaves the cube from the surface parallel to ∂x ∂y ∂z y - z plane at x = a.
Chapter 24 Electrostatics 645 Flux leaving the cube = ES 42. We have = El0x (a2) = El0a (a2) (at x = a) √R q 2 θ l + Substituting the values, we get l2 φ = (5 × 103) (10− 2)3 (2 × 10− 2) R = 2.5 × 10− 1 φ = q (1 − cos θ) 2 ε0 = 0.25 N-m2/C Here, flux due to + q and − q are in same direction. At all other four surfaces, electric lines are tangential. Hence, flux is zero. ∴ φ total = 2φ = q (1 − cosθ) ε0 ∴ φ net = (0.25) N - m2/C = qin ε0 = q 1 − l ∴ qin = (0.25) (ε0) ε0 R2 + l2 = (0.25) (8.86 × 10− 12) = 2.2 × 10− 12 C y C = (a, 0, 0) 40. See the hint of Example-1 of section solved 43. O x examples for miscellaneous examples. We have q C φ = (1 − cos θ) Q Flux passing through hemisphere = flux passing 2ε0 through circular surface of the hemisphere. But, Q = φ total For finding flux through circular surface of ε0 hemisphere we can again use the concept used in above problem. ∴ φ = (φ total ) 1− cos θ ...(i) 2 q Given that φ = 1 (φ total ) 45° 4 a 60° a b R φ = q (1 − cos θ) 2ε0 Substituting in Eq. (i), = q (1 − cos 45° ) 2ε0 we get, θ = 60° R = tan 60° = 3 = q 1 − 1 b 2ε0 2 ∴ R = 3b 41. (a) S j = (L2) (− $j) σ φ s1 = E⋅S j = − CL2 σ ∴ 44. Similarly, we can find flux from other surfaces. C Note Take area vector in outward direction of the A cube. rB (b) Total flux from any closed surface in uniform electric field is zero. R σ (4πr2 + 4πR2) = Q
646 Electricity and Magnetism ∴ σ = 4π Q R2) VC = σε0a ac − σε0b bc + σc (r2 + ε0 Potential at centre σ a2 b2 c = potential due to A + potential due to B = − + = σr + σR ε0 c c ε0 ε0 (c) WA→C = 0 = σ (r + R) ∴ VA = VC ε0 or (a − b + c) = a2 − b2 + c c = Q r + R 4πε0 r2 + R2 or a + b = c 45. P Q C2 120° C1 48. (a) Sphere –Q +Q Q Ring (Total charge = q0) σi = −Q and σ0 = Q 4 πa2 4 πa2 Charge on arc PQ of ring = q0 3 This is also the charge lying inside the closed Q sphere. ∴ φ through closed sphere = qin (b) ε0 –Q = (q0 /3) = q0 Q +q ε0 3ε0 σi = −Q 4 πa2 46. +q and σ0 = (Q + q) 4 πa2 B A –q (c) According to Gauss’s theorem, 2q E = 1 qin 4 πε 0 r2 –2q zero If outermost shell is earthed. Then, charge on For x ≤ R, qin = Q outer surface of outermost shell in this case is and r = x both cases. always zero. ∴ = 1 Q 4 πε 0 x2 47. = σa − σb + σc E ε0 ε0 ε0 (a) VA = σ (a − b + c) 49. Let Q is the charge on shell B (which comes from ε0 earth) σa σb σc VB = 0 ε0 ε0 ε0 VB = ba − + ∴ kq + kQ − kq = 0 bbc = σ a2 − b + ∴ Q = b − 1 q ε0 c c b
Chapter 24 Electrostatics 647 Charges appearing on different faces are as shown Total charge on A + C is 3q. Therefore, below. − q + q1 − q2 + q3 = 3q …(i) VB = 0 ∴ k q1 − q + k q22−Rq1 + k q3 − q2 = 0 2R 3R …(ii) q VA = VC –q Q +q ∴ k q1 − q + k q2 − q1 + k q33−Rq2 –(Q +q) R 2R Q = k q1 − q + k q2 − q1 + k q3 − q2 …(iii) 3R 3R 3R Q + q = bc q In the above equations, 50. k= 1 4 πε 0 A Solving above there equations, we can find q1, q2 O and q3. q1 52. (a) From Gauss’s theorem –q1 B q2 E = 1 qin = 1 2Q C 4πε0 r2 4πε0 r2 –q2 q3 = Q 2πε0r2 Total charge on A + C is 3q (b) According to principle of generator, potential ∴ q1 − q2 + q3 = 3q …(i) difference depends only on qin . …(ii) VB = 0 ∴ PD = 2θ 1 − 31R = Q 4 πε 0 R 3πε0R kq1 q2 − q1 q3 − q2 ∴ 2R + k 2R + k 3R = 0 (c) According to principle of generator, whole inner charge transfers to outer sphere. VA = VC (d) Vin = 0 kq1 q2 − q1 q33−Rq2 ∴ R + k 2R + k ∴ k qin − kQ = 0 ⇒ qin = Q R 3R 3 = kq1 + k q2 − q1 + k q3 − q2 …(iii) 53. (a) At r = R, 3R 3R 3R 1 Q 2Q 3Q In the above equations, k = 1 . V = 4 πε 0 R − 2R + 3R 4 πε 0 Solving these three equations, we can find the = 1 Q asked charges. 4πε0 R 51. At r = 3R V = 1 Q − 2Q + 3Q = Q 4 πε 0 3R 3R 3R 6πε0R q (b) E = 1 qin A πε0 r2 –q B q1 –q1 4 C 1 Q− 2Q −Q q2 = 4πε0 (5R / 2) = 25πε0R2 –q2 q3
648 Electricity and Magnetism Minus sign implies that this electric field is Now, qA = q = Q radially towards centre. 2 qC = 4Q − q = 7Q 2 (e) E = 1⋅ qin = 1 q− 2Q = − 3Q 4 πε 0 r2 4 πε 0 (5R / 2)2 50πε0R2 (c) U1 Minus sign indicates that electric field is U2 Q radially inwards. U3 –Q –Q LEVEL 2 +Q Single Correct Option +2Q UT = U1 +U2 + U3 1. Let us conserve angular momentum of + 2q about where, U1 = 1 Q2 the point at + Q. where, 2 C1 mv1r1 sin θ1 = mv2 r2 sin θ2 where , 1 21R (m)(v )(R ) sin 150° and R where, C1 = 4πε0 − = m v rmin sin 90° 3 Q2 U2 = 1 C2 ∴ rmin = 3R 2 2 C2 = 4πε0 1 − 31R 2. (vA)y = v ⇒ (vB )y = 2v sin 30° = v 2R Since, y -component of velocity remains U3 = 1 (2Q )2 unchanged. Hence electric field is along (− i$) 2 C3 direction. Work done by electrostatic force in C3 = 4πε0 1 − ∞1 moving from A to B = change in its kinetic energy 3R ∴ (eE) (2a − a) = 1 m (4v2 − v2) 2 C E = mv2 B 2ea (d) A or E = − 3mv2 $i 2ea q Rate of doing work done = power –q = Fv cos θ –2Q +q 2Q –q 3mv 2 2 2Q = 2ea (e) (2v) cos 30° Total charge on (A + C ) is 3Q + Q = 4Q. = 3 3 mv2 2a Now, VA = VC E ∴ kq − k (2Q ) + k (4Q − q) 3. R 2R 3R = k (Q − 2Q + 3Q ) …(i) a bx 3R +q –q Solving this equation, we get Just to the right of a, electric field is along ab q= Q (∴positive) and tending to infinite. Similarly, 2
Chapter 24 Electrostatics 649 electric field just to the left of b electric field is Let Q charge comes on shell-C from earth. Then, again along ab (∴positive) and tending to infinite. VC = 0 Q –Q ∴ kq + kQ − kq = 0 3a 3a 4a E1 Solving, we get 60° 4. Q 60° E2 –Q Q=− q 4 Q E3 5 Now, VA = kq − kq/ 4 − kq = kq 2a 3a 4a 6a –Q and VC = 0 1 P ∴ VA − VC = kq 4πε0 x3 6a E = E1 = E2 = E3 = q 8. ρ= (4 /3) πR3 = 4 1 2Qa ∴ q = 4 πρR3 πε0 x3 3 Resultant of E1 and E3 is also equal to E along E2 3 1 q 1 ⋅q 2 πε R 4πε0 R ∴ Enet = 2E (along E2) VC − VS = 4 ⋅ − = Qa 0 πε 0 x 3 =q 5. From centre to the surface of inner shell, potential 8πε0R will remain constant = 10 V (given). Substituting the value of q , we have 6. By closing the switch whole inner charge transfers VC − VS = ρR3 6ε0 to outer shell. F2 A F1 9. U2 –q q ⇐ U2 q B vA = 0 U1 q vC = 0 Heat produced = U i − U f C = (U 1 + U 2) − U 2 Hence, in between A and C there is a point B, where speed of the particle should be maximum. =U1 = 1 q2 2 C F1 = mg = constant F2 = electrostatic repulsion (which increases as the where, C = 4 πε 0 = 8πε0a particle moves down) (1/a − 1/2)a From A to B kinetic energy of the particle ∴ Heat = q2 = kq2 increases and potential energy decreases. Then, 16πε0a 4a from B to C kinetic energy decreases and potential energy increases. 7. –q 10. Over Q1, potential is + α. Hence, Q1 is positive. q VA = 0 and A point is nearer to Q2. Therefore, Q2 A should be negative and |Q1 | > |Q2|. B At A and B, potential is zero, not the force. C Equilibrium at C will depend on the nature of D charge which is kept at C.
650 Electricity and Magnetism 11. V1 is positive and V2 is negative. Hence at all kmin + qVC1 = 0 + qVC2 …(i) points, VC1 = 1 Q − Q 4 πε 0 2R R V1 > V2 VC2 = 1 Q − 2QR 12. Just to the right of q1, electric field is + α or in 4 πε 0 R positive direction (away from q1). Hence, q1 is Substituting these values in Eq. (i), we can find positive. Just to the left of q2, electric field is − α or towards left (or away from q2). Hence, q2 is also K min . positive. 20. E = Ex$i + Ey$j Further E = 0 near q1. Hence, q1 < q2 Now we can use, ∫ dV = − ∫ E ⋅ dr 13. Electric lines of forces of q will not penetrate the two times and can find values of Ex and Ey. conductor. 21. Let P = (x, y) 14. E = 400 cos 45° $i + 400 sin 45°$j ∫VA − VB = − A V=0 E ⋅ dr X = 9a B where, dr = dx $i + dy $j X=a C X = 5a 15. qA will remain unchanged. Hence, according to principle of generator R = 4a potential difference will remain unchanged. VA′ − VB′ = VA − VB (as VB′ = 0 ) r`1 = (x + 3a)2 + y2 or VA′ = VA − VB 16. WT = 0 r2 = (x − 3a)2 + y2 WFe = (Fe ) (displacement in the direction of force) Vp = 1 Q − 2Q = 0 …(i) = Kinetic energy of the particle. 4 πε 0 r2 r1 ∴ 1 mv2 = qE l − l cos 60° Substituting values of r1 and r2 in Eq. (i), we can 2 2 2 see that equation is of a circle of radius 4a and ∴ v = qEl centre at 5a. m 17. L = mv r⊥ = m (at) (x0) 22. Fx = 0 = m qE0 t (x0) or L∝t ∴ ax = 0 m Fy = qE 18. U i + Ki = U f + K f ∴ ay = qE m or qVi + 1 mvm2 in = qV f + 0 x = vt and 2 y = 1 ayt 2 = 1 qmE t2 2 2 or 1 Q + 1 mvm2 in + q 3 × Q q 4πε0 R 2 2 4πε0R Substituting t = x in expression of y , we get v From here, we can find vmin . 1 qmEvx22 y = 2 Kmin 19. C2 2R –Q KE = 1 m (vx2 + v2y ) C1 2 +Q qE kC1 + UC1 = kC2 + UC2 where, vx = v and vy = ayt = m t
Chapter 24 Electrostatics 651 23. qE = mv2 30. Vp = 2 1 q 1 ⋅ q πε0 a −2 r 4 4πε0 b2 + a2 q λ = mv2 q ba22 − 1/ 2 2πε 0r r 4 πε 0a 2 − 2 1 + = ∴ v = gλ = 2kqλ Now, 2πε0m m Since, b < < a, we can apply binomial expansion T = 2πr ∴ Vp = q − 2 − b2 v 4 πε 0a 2 1 2a2 Tmin π = qb2 C 4 πε 0a3 24. αT Tmax 31. Let E = magnitude of electric field at origin due to qE charge ± q. Then, mg 5E E1 T sin α = qE T cos α = mg (i) qE 5E mg ∴ α = tan − 1 Minimum tension will be obtained at α + π. E1 = (5E)2 + (5E)2 = 5 2E 25. Energy required = ∆U = U f − U i 5E E2 = 1 q2 − q2 − q − q2 − q2 − q2 + q2 (ii) 4 πε 0 a 2a a a 2a a a 5E = q2 [ 2 + 1] E2 is again 5 2E. 4 πε 0a Similarly, we can find E3 and E4 also. 26. On both sides of the positive charge V = + ∝ just over the charge. q 2 R2 27. U = 1 ⋅ q2 (a = side of triangle) 32. U = 1 ε0E2 = 1 ε0 1 2 2 4πε0 4πε0 a W =U f − U = 3 1 q2 −U 9 × 109 ×1 × 10− 9 2 4 πε 0 9 i a = 1 ε0 = ε0 J/m3 2 (1)2 2 = 3U − U = 2U 28. U i + Ki = U f + K f 0 + 1 mv2 = 1 ⋅ Qq + 0 1 B 2 4πε0 r v2 or r ∝ 33. If v is doubled, the minimum distance r will A remain 1 th. 4 29. See the hint of Sample Example 24.9 They have a common potential in the beginning. This implies that only B has the charge in the K = ρ = 1.6 = 2 beginning. ρ − σ 1.6 − 0.8
652 Electricity and Magnetism ∴ V = kqB or kqB = Vb Since |QB | > QA, electric field outside sphere B is b inwards (say negative). From A to B enclosed Now, suppose qA charge is given to A. Then, VA = kqA + kqB =0 charge is positive. Hence, electic field is radially a b outwards (positive). or kqA = − a kqB = − aV 38. E = − ∂V i$ + ∂V $j = − [(− ky) i$ + (− kx) $j] b ∂x ∂y Now, VB = kqA + kqB ∴ |E| = (ky)2 + (kx)2 b b =− aV +V =V 1 − ba = k x2 + y2 = kr b ∴ |E| ∝ r 34. Let E = Ex $i + Ey$j + Ezk$ More than One Correct Options Apply ∫ dV = − ∫ E⋅ dv 1. (a) VA = 2V = kqA + kqB R 2R three times and find values of Ex , Ey and Ez. Then, again apply the same equation for given point. 3V kqA kqB 2 2R 2R A q′ VB = = + 35. B Solving these two equations, we get qA = 1 q qB 2 2 Let charge on B is q′ qA –qA = qB VB = 0 (b) ∴ (k (q/2)) + kq′ = 0 dr ∴ q′ = − qr 2d +q – + q′A = qA = − 1 – + q′B − qA 36. + – + (c) & (d) Potential difference between A and B P will remain unchanged as by earthing B, – + charge on will not changed. – – The induced charges on conducting sphere due to ∴ VA′ − VB′ = VA − VB + q charge at P are as shown in figure. = 2V − 3V = V 22 Now, net charge inside the closed dotted surface is negative. Hence, according to Gauss’s theorem net ∴ VA′ = V flux is zero. 2 as VB′ = 0 37. 2. T = 2uy = 2 × 10 = 2 s E=0 g 10 +QA –QB H = u2y = (10)2 = 5 m 2g 20 R = 1 axT 2 = 1 qmE T 2 2 2
Chapter 24 Electrostatics 653 1 10− 3 × 104 qE = 2 2 (2)2 = 10 m …(i) T1 …(ii) T2 3. 100 = 1 ⋅ q qE 4πε0 (R + 0.05) 75 = 1 ⋅ q If we displace the rod, τ1 = τ2 or τnet = 0 in displaced position too. Hence, equilibrium is 4πε0 (R + 0.1) neutral. Solving these equations, we get 10. Along the line AB, charge q is at unstable q = 5 × 10− 9 C 3 equilibrium position at B (When displaced from B along AB, net force on it is away from B, whereas and R = 0.1 m force at B is zero). Hence, potential energy at B is (a) V = 1 ⋅ q maximum. Along CD equilibrium of q is stable. Hence, 4πε0 R potential energy at B is minimum along CD. (9 × 109 ) 5 × 10− 9 Comprehension Based Questions 3 1. Vouter = 0 = 0.1 ∴ kQ + kQ1 = 0 2r 2r = 150 V ∴ Q1 = − Q = charge on outer shell (c) E= 1⋅ q =V = 150 4 πε 0 R2 R 0.1 = 1500 V/m (d) Vcentre = 1.5 Vsurface 2. Vinner = 0 5. Electric field at any point depends on both charges ∴ kQ2 + kQ1 = 0 r 2r Q1 and Q2. But electric flux passing from any closed surface depends on the charged enclosed by that closed surface only. 6. Flux from any closed surface = qin , ∴ Q2 = − Q1 = Q = charge on inner shell 2 2 ε0 qin = 0, due to a dipole. Charge flown through S2 = initial charge on inner shell − final charge on it 8. E = k qin k = 1 r2 Here, 4πε0 Q = Q − Q2 = 2 ∴ EA = Ec = 0 3. After two steps charge on inner shell remains Q or but, EB ≠ 0 2 V = kq (r ≤ R) half. R So, after n-times V = kq r (r ≥ R) qin = Q (2)n 9. Fe = qE Now, according to the principle of generator, potential difference depends on the inner charge only. Hinge force ∴ PD = qin 1 − 1 4 πε 0 r 2r Higher force = 2qE Fe = qE = 1 1 Q 2n + 4πε0 r (towards left)
654 Electricity and Magnetism 4. According to Gauss’s theorem, q2 –Q E = 1 qin …(i) 4πε0 r2 For r ≤ R ∫qin = r (4πr2)⋅ dr ⋅ρ (ii) 0 ∫= r (4 πr2) (ρ0 ) 1 − Rr dr Vinner = 0 when solid sphere is earthed 0 Q kq2 − kQ = 0 = 4 πρ0 r3 − r4 ab 3 4R Q q2 = Q ba Substituting in Eq. (i), we get E = ρ0 r − r2 10. Whole inner charge transfers to shell. ε 3 4 r ∴ Total charge on shell = q2 − Q a 1 5. For outside the ball, = Q b − E= 1 qtotal …(i) Match the Columns 4 πε 0 r2 where, ∫qtotal = R (4 πr2 ) (ρ0) 1 − Rr dr 1. (a) EC and EF are cancelled. EE and ED at 60° 0 (b) EB and EE are cancelled. EF and ED at 120°. Substituting this value in Eq. (i), we get (c) EB and EE are cancelled. Similar, EF and EC are cancelled. E = ρ0R3 12 εr2 (d) EF and ED at 120°. So, their resultant is E in the direction of EE. Hence, net is 2E. 6. For outside the ball, electric field will 2. ∫ dV = − ∫ E ⋅ dr continuously decrease. Hence, it will be maximum somewhere inside the A ball. For maximum value, VA − VB = − E ⋅ dr ∫∴ dE = 0 B dr 3. V = kq ⇒ kq = VR R ∴ d ρε0 r − r2 = 0 (a) V = kq (1.5R2 − 0.5r2) dr 3 R3 4R Solving, we get r = 2R = VR 3 R 2 − 1 R2 2 3 R3 2 2 7. Submitting r = 2R in the same expression of = 11 V 3 8 electric field, we get its maximum value. (b) V = kq = VR = V r 2R 2 8. Potential difference in such situation depends on (c) E = kq ⋅r inner charge only. So, potential difference will R3 remain unchanged. Hence, (VR ) R2 V V ∆V = Va − Vb (R3) 2R 2 q1 = 0 = = = (if R = 1 m) Q –Q 9. (d) E = kq = VR r2 (2R)2 (i) = V = V for R = 1 m 4R 4
Chapter 24 Electrostatics 655 Subjective Questions q3 = − q2 = + 2 Q q4 = 2Q − q3 = 0 1. (a) By comparing this problem with spring-block q5 = − q4 = 0 system problem suspended vertically. Here, mg ≡ qE = 50 × 10−6 × 5 × 105 = 25 N 12 34 56 X max = 2 mg/K Here, X max = 2 qE / K = 2 × 25 = 0.5 m 100 or = 50 cm Ans. (b) Equilibrium position will be at x = mg/K. Here, it will be at x = qE/ K = 25 = 0.25 m Q 2Q 3Q 100 O x 2q or 25 cm h Ans. q (c) Force QE is constant force, which does not 4. w affect the period of oscillation of SHM. h ∴ T = 2π m = 2π 4 QQ K 100 = 2π s = 1.26 s Ans. (a) Net torque on the rod about O = 0 5 1 Q (2q) L2 w L x (d) µmg = 0.2 × 4 × 10 = 8 N πε . h2 + 2 − 4 Therefore, here constant force will be 0 qE − µmg = 25 − 8 = 17 N = F (say) = 1 . Q.q L2 πε h2 X max = 2F = 2×7 4 0 K 100 x = L + Qq Ans. = 0.34 m Ans. 2 1 (4πε0)wh2 2. Total charge on ring = λ (2πa) = q (say) (b) There will be no force from the bearing it, Electric field at distance x from the centre of ring. w = net electrostatic repulsion from both the E = 1 qx = λax charges. = 1 Q (3q) 4πε0 . (a2 + x2)3/ 2 2ε0(a2 + x2)3/ 2 ∴ w . 4 πε 0 h2 Restoring force on − Q charge in this position or h = 3Qq Ans. would be 4 πε 0w F = − QE = − λaQx ∂V ∂V $j 2ε0(a2 + x2)3/ 2 ∂x ∂y 5. E = − i$ + = (− 3i$ + 4$j) N/C For x << a, F = − λaQ x = − λQ x a = qE m 2ε0a3 2ε0a2 = 10−6 (− 3$i − 4$j) Comparing with F = − kx, 10 k = λQ = (− 3 × 10−7 i$ − 4 × 10−7 $j) m/s2 2ε0a2 T = 2π m = 2π 2ε0ma2 Ans. When particle crosses x-axis, y = 0. k λQ Initial y- coordinate was 3.2 m. and ay = − 4 × 10−7 m/s2 qnet 3. q1 = q6 = 2 = 3Q 2 × 3.2 4 × 10−7 q2 = Q − q1 = − 2 Q ∴ y = 0 at time t = = 4000 s
656 Electricity and Magnetism At this instant x-coordinate will be Q –Q x = xi + 1 axt2 2 = 2 + 1 (− 3 × 10−7 )(4000)2 = − 0.4 m –Q Q 2 –Q Q Now, Vi = (3 × 2) + (4 × 3.2) = 18.8 V Vf = (3) (− 0.4) = − 1.2 V ∆V = 20 V ∴ Speed, v = 2q∆V Q –Q m = 2 × 10−6 × 20 with decrease in L, potential energy will decrease. 10 Therefore, cube should shrink as the conservative = 2.0 × 10–3 m/s Ans. forces act in the direction of decreasing potential 6. From work-energy theorem, energy. v C 60° Increase in KE of the system = decrease in PE qE or 8 1 mv2 =Ui −U f 2 l mg = Q2 3 6+ 2−3 3 1 − L1 60° πε0 6 nL O E l or v = Q 2(1 − n)(3 6 + 2 − 3 3) Ans. 4nmπLε0 6 u 8. Let charge q1 comes from the earth on outer shell 1 m (v2 – u2) = – mgl (1 + sin 60° ) + qE l cos 60° q1 = – Q –Q 2 Substituting the values, we get Q q1′ u2 – v2 = 32.32 …(i) ⇒ Further, at C tension in the string is zero. . Hence, mv2 = mg sin 60° – qE cos 60° …(ii) Vouter = 0 l or v2 = 3.66 Ans. 1 Q + q1 = 0 4 πε 0 2r 2r From Eqs. (i) and (ii), we get u = 6 m/s or q1 = − Q 7. There are total 28 pairs of charges. When S2 is closed and opened, 12 pairs → Q and − Q → distance L Vinner = 0 12 pairs → (Q and Q ) or (− Q and −Q )→ 2L 1 q1′ Q 4 pairs → Q and − Q → 3L ∴ 4 πε 0 r − 2r = 0 ∴ U = 12 4 1 − Q2 + 12 4 1 . Q2 or q1′ = Q πε0 πε 2 L 0 2L + 4 1 − Q2 Proceeding in the similar manner after n such π ε0 operations we get, 4 3L Charge on the inner shell, = − Q2 3 6+ 2−3 3 qn′ = Q πε0L (2)n 6
Chapter 24 Electrostatics 657 and the potential difference between the shells, In the second contact, S1 again acquires the same charge Q. ∆V = qn′ 1 − 21r 4 πε 0 r Therefore, total charge in S1 and S2 will be ∆V = 1 Q Ans. Q + q1 = Q 1 + R (2)n + 1 4πε0r R+ r 9. (a) Over charge Q2, field intensity is infinite along This charge is again distributed in the same negative x-axis. Therefore, Q2 is negative. ratio. Therefore, charge on S2 in second contact, Beyond x > (l + a), field intensity is positive. Therefore, Q1 is positive. q2 = Q 1 + R R R+ r R + r (b) At x = l + a, field intensity is zero. kQ1 kQ2 Q1= l + a 2 R R 2 (l + a)2 a2 Q2 a + R + r ∴ = or = Q R r + (c) Intensity at distance x from charge 2 would be Similarly, E = kQ1 − kQ2 R R 2 R 3 (x + l)2 x2 + r R+ r q3 = Q R + r + R + For E to be maximum dE = 0 and dx 2 n 2kQ1 2kQ2 R R r R or − (x + l)3 + x3 = 0 qn = Q + R + + … + R + r R + r 1 + l 3 Q1 l + a 2 x Q2 a or = = R R n r 1 − R + r or qn = Q …(ii) or 1+ l = l + a 2/ 3 a (1 − rn) x a Sn = l (1 − r) or x = a + 2/ 3 l a − 1 Therefore, electrostatic energy of S2 after n such contacts or b = + l Ans. Un = an2 = qn2 or Un = qn2 a a 2/ 3 2C 2(4 πε 0R ) 8πε0R l − 1 where, qn can be written from Eq. (ii). 10. Capacities of conducting spheres are in the ratio of QR R R n − 1 R+r 1 + R+ R + r their radii. Let C1 and C2 be the capacities of S1 (b) qn = r +…+…+ and S2, then as n → ∞ C2 = R C1 r q∞ = QR 1 R+r R (a) Charges are distributed in the ratio of their 1 − r capacities. Let in the first contact, charge R+ acquired by S2, is q1. Therefore, charge on S1 will be Q − q1. Say it is q1′ . = QR R + r = Q R S ∞ = a R+r r r 1− r ∴ q1 = q1 = C2 = R q1′ Q − q1 C1 r q∞2 Q 2R2 / r2 2C 8πε0R It implies that Q charge is to be distributed in ∴ U∞ = = S2 and S1 in the ratio of R/r. Q 2R 8πε0r2 ∴ q1 = Q R …(i) or U∞ = Ans. + r R
658 Electricity and Magnetism 11. vx = v = 2qV , t = l = l m 14. From energy conservation principle, m vx 2qV Ki + Ui = Kf + U f ay = qE = qat = dv y or 1 mv2 + (+ q)Vi = 0 + (+ q)V f m m dt 2 – or v= 2q (− Vi + Vf ) m +v y = 2q −Q + 8Q + Q − 8Q 1 E x m 10R 5R R 4R 4 πε 0 + = Qq 3 10 − 5 Ans. Integrating both sides, we get 2πε0mR 5 10 vy = qat2 15. (a) E = (R2 kQx , 2m + x2)3/ 2 or vy = 2qma l2 m = al2 (R 2 + x2)3/ 2 − x. 3 (R2 + x2 )1/ 2(2x) 2qV 4V 2 dE = kQ dx (R2 + x2)3 Now, angle of deviation /θ vy al2 2qV = tan −1 vx = tan −1 dE R2 + x2 − 3x2 4V dx m or = kQ (R2 + x2)5/ 2 al2 m = tan −1 Ans. R2 − 2x2 4V 2eV Q or dE = 4 πε 0 (R 2 + x2 )5/ 2 dx 12. From energy conservation, R2 − 2x2 UC + KC = U D + KD ∴ |∆E| = Q ∆x Ans. 4 πε 0 (R2 + 2 )5/ 2 x 9 × 109 × (5 × 10−5)( − 5 × 10−5) or 2 5 + 4 Here, ∆x = 2a Qqa R2 − 2x2 9 × 109 × (5 × 10−5)(− 5 × 10−5) ∴ F = |q∆E| = 2πε0 = AD + (R2 + x 2 )5/ 2 2 0 Solving we get AD = 9 m (b) W = U f − U i ∴ Maximum distance, = − pE cos 180° + pE cos 0° = 2pE OD = (9)2 − (3)2 = 2(q)(2a) 4 1 (R2 Qx πε0 + x2)3/ 2 = 72 m Ans. aqQx 13. From conservation of energy, = πε0(R2 + x2)3/ 2 Ans. Ui + Ki =U f + Kf 16. From conservation of mechanical energy and or qVi + Ki = qV f + K f conservation of angular momentum about point O, we have or q Q + 1 mv2 4 πε0r 2 v1 Q R 2 O r1 πε0R 3 1.5R +Q = q 2 − 0.5 + 0 r2 4 4 B A or 1 mv2 = 11 Qq − Qq v2 2 3πε0R 4πε0r or v= Qq r −R + 38 Ans. 1 mv12 − 1 . Qq = 1 mv22 − 1 . Q.q …(i) 2πε0mR r 2 4πε0 r1 2 4πε0 r2
Chapter 24 Electrostatics 659 and mv1r1 sin 90° = mv2r2 sin 90° ∴ E = 1 . q1 or v1r1 = v2r2 4 πε 0 r2 Solving these two equations, we have …(ii) = 9 × 109 × (200/9) × 10−12 (1.25 × 10−2)2 Qqr2 v1 = 2πε0mr1(r1 + r2) = 1.28 × 103 V/m Ans. Qqr1 18. (a) Fnet = 2F cos θ 2πε 0mr2 (r1 and + r2 ) = v2 Ans. = 2kQ.q . x0 1 ( R2 + x02 )2 R2 + x02 4 πε 0 Here, k = 17. Let q1 : q2 and q3 be the respective charges. Then, 2kQqx0 (R2 + x02)3/ 2 = +Q 2+ 2 0 q1 q2 q3 √R x RF x0 θ θ –q F 10 = 9 × 109 q1 + q2 + q3 10−2 1 2 4 +Q 0= 9 × 109 q1 + q2 + q3 We can generalised the force by putting x0 = x, 10−2 2 2 4 we have and 40 = 9 × 109 q1 + q2 + q3 F = − 2kQqx Ans. 10−2 4 4 4 (R2 + x2)3/ 2 Solving these equations, we get (b) Motion of bead will be periodic between 200 × 10−12 200 × 10−12 x = ± x0 Ans. 9 q1 = + C, q2 = − C and (c) For x << 1, R2 + x2 ≈ R2 R q3 = 3200 × 10−12 C 9 = − 2kRQ3 q F = − 2mkRQ3q or F x or a= m x (a) At r = 1.25 cm 9 × 109 (200/9) × 10−12 − 200 × 10−12 Since a ∝ − x, motion will be simple harmonic 10−2 1.25 2 in nature. + (3200/9) × 10−12 V = 4 Comparing with a = − ω2x, ω = 2kQq mR3 =6V Ans. x = x0 cos ωt (as the particle starts from extreme position) (b) Potential at r = 2.5 cm v = dx = − ω x0 sin ωt Ans. (200/ 9) × 10−12 − 200 × 10−12 dt 2.5 + (3200/9) V = 9 × 109 4 2.5 (d) Velocity will become zero at t = T /2 = π/ω 10−2 × 10−12 v=0 v=0 = 16 V Ans. x = – x0 x = x0 (c) Electric field at r = 1.25 cm will be due to or t = π mR3 Ans. charge q1 only. 2kQq
660 Electricity and Magnetism 19. Net charge between r = a to r = r would be 21. Two forces will act on the tank. ∫ ∫Q = r ρ (4πr2) dr = r C (4πr2) dr (a) Electrostatic force, a ar Q = 2πC (r2 − a2) (b) Thrust force. Let v be the velocity at any instant. Then, k (Q + q) k [2πC (r2 − a2 ) + q] Fnet = QE − mnv r2 r2 Er = = or (m0 + mnt) dv = QE − mnv dt From this expression, we can see that it we put v dv t dt = q ∫ ∫or C = 2πa2 0 QE − mnv 0 m0 + mnt = kq = or ln QE = ln m0 + mnt a2 QE − mnv m0 Er constant Ans. 20. dF = λ (Rdθ)E0 or QE = m0 + mnt QE − mnv m0 ++ + + dF t + mnt + or v = QE m0 Ans. + +θ 22. qE = 30 N, vertical component of electric force – = 30 sin 30° = 15 N and horizontal component of – electric force = 30 cos 30° = 15 3 N – dF ––––– ay = mg − 15 = 30 − 15 = 5 m/s2 (downwards) m 3 f Perpendicular distance between two equal and 15 3 3 opposite pairs of dF will be ax = = 5 3 m/s2 r⊥ = 2R sin θ T1 = 2uy = 2 × 20 sin 30° = 4 s ∴ dτ = dFr⊥ = 2λR2E0 sin θdθ ay 5 ∫∴ τ= π /2 = 2λR2E0 (clockwise) T2 = eT1 = 2 s Horizontal velocity after first drop dτ = (20 cos 30° ) + axT1 0 = (10 3) + (5 3)4 These pairs of forces will not provide net force. Let force of friction on ring is f in forward direction. For pure rolling to take place, = 30 3 m/s a = Rα ∴ Horizontal distance travelled between first drop and second drop or f = R τ − fR m mR2 1 = (30 3 )T2 + 2 axT22 or f = τ − f R = (30 3)(2) + 1 (5 3)(2)2 2 or f = τ = λRE0 Ans. 2R = 70 3 m Ans.
25. Capacitors INTRODUCTORY EXERCISE 25.1 ∴ A = Cd (c) ε0 1. U = 1 q2 3. (a) σ= q 2C A ∴ [C ] = q2 = A2 T2 K = E0 = 3.20 × 105 U ML2 T−2 E 2.50 × 105 = [M − 1L−2 T4A2 ] = 1.28 2. Charge does not flow if their potentials are same. (b) σi = σ0 1 − 1 K 3. q1 = C1V1 = 10 µC (E0 ε0)1 1 q2 = C2V2 = − 40 µC = − K (a) V = qTotal = −30 µC = − 10 volt = (3.20 × 105) (8.86 × 10− 12) 1 − 1.128 CTotal 3 µF = 6.2 × 10−7 C/ m2 (b) q1′ = C1V and q2′ = C2V (c) ∆U = C 1C 2 (V1 − V2)2 2(C1 + C2) INTRODUCTORY EXERCISE 25.3 INTRODUCTORY EXERCISE 25.2 1. Cnet = 2 µF, q = CV = 2 × 15 = 30 µC 1. q = CV Now, this q will be distributed between 4 µF and 2. (a) V = q 2 µF in direct ratio of their capacities. C 2. Cnet = 3 µF, q = CV = 3 × 40 = 120 µC. (b) C = ε0A Now, this q will be distributed between 9 µF and d 3 µF in the direct ratio of their capacities. Exercises LEVEL 1 5. Charges are same, if initially the capacitors are Assertion and Reason uncharged. 1. Capacitance of conductor depends on the Further, V=q C dimension of the conductor and the medium in which this conductor is kept. Hence, V∝1 C 3. Energy supplied by the battery is if q is same. ∆qV = (CV )(V ) = CV 2. 6. Charge (or current) will not flow in the circuit as Energy stored in the capacitor is 1CV 2. 2 they have already the same potential, which is a condition of parallel grouping. 4. In graph-1, discharging is slow. Further, Hence, τC1 > τC2 q1 = C1V = C1 = 1 Further, q2 C2V C2 2 τC = CR 7. Capacitor and R2 are short-circuited. Hence, ∴ τC ∝ R (as C = constant) current through R2 is zero and capacitor is not charged.
662 Electricity and Magnetism 8. Capacitor and resistance in its own wire are 9. At t = 0, when capacitor is under charged, directly connected with the battery. Hence, time equivalent resistance of capacitor = 0 constant during charging is CR. In this case, 6 Ω and 3 Ω are parallel (equivalent = 2Ω) 9. U = 1 q2 or U ∝ 1 as q is same in capacitors ∴ Rnet (1 + 2) Ω = 3 Ω ∴ Current from battery = 12 = 4A 2C C (if initially they are uncharged) 3 = Current through 1Ω resistor 10. By inserting dielectric slab, value of C2 will 10. Final potential difference = E increase. In series, potential difference distributes in inverse ratio of capacitance. If capacitance C2 is ∴ Final charge = EC increased PD across C2 will decrease. If C2 is increased, charge on capacitors will also increase. i/4 a b So, positive charge or current flows in clockwise i/4 direction. Objective Questions 11. 1. F = qE = q σ = q q q will not change. i 3i/4 i 2ε0 2 Aε 0 ∴ F = constant 3. C = C1 + C2 E = (4πε0 a) + (4πε0 b) Va − i R − iR = Vb = 4πε0 (a + b) 4 4. Vnet = V1 + V2 + … (in series) ∴ Va − Vb = 5 iR = 10 …(i) 4 =V +V +… = nV Rnet = R + (3R) (R) (3R + R) 5. V32 = V5 = 6 V ∴ q5 = CV = 30 µC = 7R 4 3 × 22 q32 = 3 + × 6= 7.2 µC ∴ i = E = 47RE (7/ 4 )R ∴ q5 = 30 q2 7.2 Substituting in Eq. (i), we have A 45 47RE × R = 10 6. 60 V 15 V 30 V ∴ E = 14 V B 15 V 12. All capacitors have equal capacitance. Hence, 7. Q qE = mg equal potential drop (= 2.5 V) will take place across all capacitors. +σ q V = 4 πr3ρ g VN − VB = 2.5 V 8. d 3 0 − VB = 2.5 V VB = − 2.5 V E=0 V ∝ r3 ∴ q Further, VA − VN = 3 (2.5) V = 7.5 V –σ ⇒ E = σ = constant ∴ VA = + 7.5 V (as VN = 0) ε0 E=0 13. q = CV = 200 µC E0 In parallel, the common potential is given by
Chapter 25 Capacitors 663 V = Total charge 20. V = Ed = σ d Total capacity 2ε0 = 200 µC = 50 V ∴ d = 2ε0V (2 + 2) µF σ Heat loss = U i − U f = 2 (8.86 × 10− 12 ) (5) = 1 (2 × 10− 6) (100)2 − 1 (4 × 10− 6) (50)2 10− 7 22 = 0.88 × 10− 3m = 5 × 10− 3J = 0.88 mm = 5 mJ 21. Three capacitors (consisting of two loops are 14. P = i2R = (i0e− t/ η)2 R short-circuited). = (i02R) e− 2t/ η 22. The equivalent circuit is as shown below. = P0e− t/ (η/ 2) Hence, the time constant is η. 1 µF 1 µF 2 15. Common potential in parallel grouping X 1 µF Y = Total charge 2 µF Total capacity = EC = E 23. C1 = CRHS + CLHS 2C 2 16. VA − 6 − 3 × 2 + 9 − 3 × 3 = VB = K2 ε0(A/2) + K1ε0(A/2) 1 dd ∴ VA − VB = 12 V = ε0A (K 1 + K2) = 5ε 0 A 2d 2d 17. In steady state condition, current flows from C2 = ε0A d/2 d/2 outermost loop. d/2 + i = 12 = 1.5 A d − d/2 − + 6+ 2 K1 K2 Now, VC = V6Ω = iR = 2ε 0 A K 1K 2 = 1.5 × 6 = 9 V d K1 +K 2 ∴ q = CVC = 18 µC = 12 ε0A 5d 19. Horizontal range, …(i) ∴ C1 = 25 R = 2ux × uy = l C2 24 g Maximum height, H = u2y = d 24. A balanced Wheatstone bridge is parallel with C. 2g …(ii) 25. First three circuits are balanced Wheatstone bridge Dividing Eq. (ii) by Eq. (i), we have circuits. 1 uy = d 26. C = CLHS + CRHS 4 ux l = K1ε0 (A/2) + ε0 (A/ 2) or uy = u sin θ d d/2 − d/2 + d/2 d/2 ux u cos θ d − + K2 K3 = tan θ = 4d = ε0A K1 + K2K3 l d 2 K2 + K3
664 Electricity and Magnetism C V2 V3 = σ d = 2.5 q d C V2 ε0 ε0A C 27. V1 C/2 B Capacitance, C = ε0A d V1 C V4 A V1 3. All three capacitors are in parallel with the battery. V3 PD across each of them is 10 V. So, apply q = CV for all of them. C = ε0A = 7 µF 4. Capacitor and resistor both are in parallel with the d battery. PD across capacitor is 10 V. Now, apply The equivalent circuit is as shown in figure. q = CV . V1 C V3 V2 B 5. In steady state, current flows in lower loop of the A V3 C V2 circuit. V1 V1 V4 V4 V3 i = 30 = 3A 6+ 4 C C/2 Now, potential difference across capacitor = potential difference across 4 Ω resistance. V1 C V2 = iR = (3) (4) =12 V C AB = 11C = 11 (7 µF) ∴ q = CV = (2 µF) (12 V) 7 7 = 24 µC = 11µF 6. (a) C net = C1 C2 = 2 µF C1 + C2 3 Subjective Questions qnet = C netV 1. Charge on outermost surfaces = 2 µF (1200 V) = qtotal = (10 − 4) µC 3 22 = 800 µC = 3 µC Hence, charges are as shown below. In series, q remains same. ∴ q1 = q2 = 800 µC 3 µC 7 µC –7 µC 3 µC V1 = q1 = 800 V C1 and V2 = q2 = 400 V C2 2. Charge on outermost surfaces (b) Now, total charge will become 1600 µC. This = qtotal = 2q − 3q = − q will now distribute in direct ratio of capacity. 22 2 ∴ q1 = C1 = 1 q2 C2 2 Hence, charge on different faces are as shown below. 31 1600 3 q1 = (1600) = µC 2.5q –q –q 2 –2.5q 2 q2 = 23 (1600) = 32300 µC Electric field and hence potential difference They will have a common potential (in between the two plates is due to ± 2.5 q. parallel) given by PD = Ed V = Total charge Total capacity
Chapter 25 Capacitors 665 = 1600 µC q 3 µF EC = 1600 V (EC – q0) 3 q0 7. Charge, q = CV = 104 µC t In parallel, common potential is given by ∴ q = q0 + (EC − q0) (1 − e− t/ τ C ) V = Total charge Total capacity = EC (1 − e− t/ τ C ) + q0e− t/ τ C 20 = (104 µC) Here, τC = CR (C + 100) µC 12. (a) Immediately after the switch is closed whole Solving this equation, we get C = 400 µF current passes through C1. ∴ i = E/R1 8. Charge supplied by the battery, (b) Long after switch is closed no current will pass through C1 and C2. q = CV ∴ i= E Energy supplied by the battery, R1 + R3 E = qV = CV 2 13. (a) At t = 0 equivalent resistance of capacitor is Energy stored in the capacitor, zero. R1 and R2 are in parallel across the battery U = 1CV 2 PD across each is E. 2 ∴ iR1 = E/R1 ∴ Energy dissipated across R in the form of heat iR2 = E/R2 = E − U = 1 CV 2 = U (b) In steady state, no current flow through 2 capacitor wire. PD across R1 is E. ∴ iR1 = E/R1 and iR2 = 0 9. i = i0 e− t/ τ C (c) In steady state, potential difference across Putting i = i0 , we get capacitor is E. 2 ∴ U = 1 CV 2 = 1 CE2 t = (ln 2) τC = (0.693) τC 22 10. Both capacitors have equal capacitance. Hence, (d) When switch is opened, capacitor is discharged through resistors R1 and R2. half-half charge distribute over both the capacitors. τC = CRnet = C (R1 + R2) q1 = q2 = q0 2 14. (a) Simple circuit is as shown below. q1 decreases exponentially from q0 to q0 while q2 2 increases exponentially from 0 to q0. 2 Corresponding graphs and equation are given in the answer. Time constant of two exponential equations AB will be (b) The simple circuit is as shown below. τC = (Cnet ) AB R = C R = CR 2 2 11. qi = q0 qf = EC Now, charge on capacitor changes from qi to qf exponentially.
666 Electricity and Magnetism (c) Let CAB = x. Then, or VA − VB = 3q − 10 = 5 2 2C A ∴ q = 10 µC Cx Now, V = q across each capacitor. C B 20. See the answer. Now, CAB = C + (2C ) (x) 21. In series, potential difference distributes in inverse or 2C + x ratio of capacitance. x = C + 2Cx ∴ VA = CB = C2 = 60 = 3 2C + x VB CA C1 40 2 Solving this equation, we get ∴ C2 = 1.5 C1 …(i) x = 2C Now, VA′ = C′B VB′ C′A 15. (a) V = 660 V across each capacitor or 10 = C2 Now, q = CV for both 90 (C1 + 2) (b) qnet = q1 + q2 or C1 + 2 = 9C2 …(ii) = (3.96 − 2.64) × 103 C = 1.32 × 10− 3C Solving Eqs. (i) and (ii), we get Now, common potential C1 = 0.16 µF C2 = 0.24 µF V = Total charge = 13.2 × 10− 3 and Total capacity 10 × 10− 6 22. (a) q = CV = 132 V (b) C = ε0A or C ∝ 1 dd Now, apply q = CV for both capacitors. 16. u = 1 ε0 E2 If d is doubled, C will remain half. Hence, q 2 will also remain half. 1 V 2 (c) q = CV = εd0A V = ε0 (πR3)V 2 d d = ε0 17. d = Vmax or q ∝ R2 Emax R is doubled. Hence, q will become four times or 480 µC. C = Kε0A d 23. Energy lost = energy stored = 1CV 2 ∴ A = dC = (Vmax ) (C ) 2 K ε0 (K ) (Emax ) (ε0) 24. (a) C = ε0A 1 2 d 18. 0.1 = (C 1 + C2) (2)2 …(i) (b) q = CV 1.6 × 10− 2 = 1 C 1C 2 (2)2 …(ii) (c) E = V 2 C1 +C d 2 25. (a) 1 =1+ 1+ 1 Cnet 8.4 8.2 4.2 Solving these two equations, we can find C1 and C2. ∴ Cnet = 2.09 µF 19. A B qnet = C netV = (2.09) (36) +– 10 V +– q q q shown in figure is in µC. = 75.14 µC ≈ 76 µC Now, VA − q + 10 − q = VB In series, charge remains same in all 1 2 capacitors.
Chapter 25 Capacitors 667 (b) U total = 1 C netV 2 q34 = 1 (5.5 × 10− 4) = 1.8 × 10− 4 C 2 3 (c) qtotal = (3) (76) µ C = 288 µC For finding PD across any capacitor, use the Now common potential in parallel, equation V = qtotal = 228 µC C= q V Ctotal (8.4 + 8.2 + 4.2) µF (d) U total = 1 C net V 2 28. In series, potential difference distributes in inverse 2 ratio of capacitance. q1 5V ∴ VA = C2 or 130 = C2 VB C1 100 C1 +– C2 26. q1 + 3 µF – +– or C1 = 1.3 = CA q2 – q3 4 µF+ q4 5V 6 µF + 2 µF q1 – K is made 2.5 times. Therefore, C1 will also become 2.5 times. – 10 V + C ′1 = 2.5 C1 = 2.5 C2 q4– q3 1.3 q4– q3 If we see the charge on positive plate of 6 µF or C1′ = 25 capacitor, then C2 13 q2 = − q1 − (q4 − q3) …(i) Now, VA′ = C2 = 13 VB′ C1′ 25 Now, applying three loop equations, we have 5 − q1 + q2 = 0 …(ii) or VA′ = 13 (230) = 78.68 V 36 13 + 25 10 − q2 − q3 = 0 …(iii) VB′ = 230 − VB′ = 151.32 V 62 2 × 3 5 − q3 − q4 = 0 …(iv) 29. C 23 = 2 + 3 = 1.2 µF 24 Solving these four equations, we can find q1, q2, q3 qtotal = C1V = 110 µC and q4. Common potential in parallel is given by 27. (a) Simple series and parallel grouping of V = Total charge Total capacity capacitors. = 110 = 50V (b) qnet = CnetV 1 + 1.2 = (2.5 × 10− 6) (220) = 5.5 × 10− 4 C q23 = (C23)V = 60 µC So, this much charge flows through the switch. C1, C5 and equivalent of other three capacitors are in series. Hence, charges across them are 30. (a) Simple circuit is as shown below. same. 4 µF 2 µF C2 = 2 µF C34 3 µF C3 4 µF 2.1 µF 4.2 µF C2 20 V C1 = 3 µF qtotal will distribute between C2 and C34 in 6 µF, V1 direct ratio of capacitance. ⇒ 20 V ∴ q2 = 4.2 = 2 6 µF, V2 q34 2.1 1 ∴ q2 = 2 (5.5 × 10− 4) = 3.7 × 10− 4C 3
668 Electricity and Magnetism (b) qnet = (Cnet )V = (34µF) (20V) = 60 µC 32. qtotal = C1V0 (c) Upper network and lower network both have After switch is thrown towards right, C23 and C1 same capacitance = 6 µF are in parallel. The common potential is V1 − V2 = 20 = 10V V = Total charge = C 1V0 2 VC1 = 10V Total capacity C1 + C 2C3 qC1 = (C1)(VC1 ) = 30 µC C2 +C ∴ ∴ qC2 = (C2)(VC2 ) = 20 µC 3 ∴ qC3 = (C3)(VC3 ) = 20 µC (d) VC2 = 10V, Now, qC1 = C1V = C 12V0 (e) VC3 = 5V, C1 C 2C 3 C1 C3 + C2 +C 3 31. (a) C2 C4 This is the same result as given in the answer. qC2 = qC3 = C23V C 2C 3 C1V0 C2 +C = C1 + C 2C 3 C2 +C 12 V 3 3 C 13 = 1× 3 = 3 µF 33. (a) q = Ci V = εd0A V 1+ 3 4 C 24 = 2× 4 = 4 µF Vf =q = (ε0AV /d) = 2V 2+ 4 3 Cf (ε 0 A / 2d ) VC1C3 = VC2 C4 = 12 V 3 (b) Ui = 1 2 = 1 ε 0A V 2 4 2CiV 2 d ∴ q1 = q3 = (C 13 ) (VC1C3 ) = × 12 = 9 µC q2 = q4 = (C 24 ) (VC2 C4 ) = 4 × 12 = 16 µC Uf = 1 C f V 2 = 1 ε20dA (2V )2 3 2 f 2 C1 C3 = ε0dA V 2 (b) C2 C4 (c) W =U f −Ui = 1 ε0dA V2 V2 2 V1 34. (a) After long time, capacitor gets fully charged by E1. 12 V ∴ iC = 0 = 1× 2 = 2 µF and iR1 = iR2 = E1 1+ 2 3 R1 + R2 C 12 3× 4 12 = 20 3+ 4 7 20 × 103 C 34 = = µF V1 = C34 = 12/7 = 18 = 10− 3 A = 1 mA V2 C12 2/3 7 ∴ V1 = 1285 (12) = 8.64 V (b) In steady state (with E1). VC = VR2 = VR1 = E1 = 10 V 2 V2 = 12 − 8.64 = 3.36 V Now, when the switch is shifted to position B, Now, we can apply q = CV for finding charge on capacitor (at t = 0) behaves like a battery of different capacitors. 10 V.
Chapter 25 Capacitors 669 The circuit in that case is as shown below. (d) Initially, R1 R2 3 mA + 18 V 10 V 1 mA 10 V q –p +r q – 2 mA Now, with the help of Kirchhoff’s laws we can q = CnetV = (2 µF) (18) find different currents. Final currents are = 36 µC shown in the diagram. Finally, 35. (a) Va = 18 Vand Vb = 0 as no current flow through + the resistors. q1 12 V ∴ Va − Vb = 10 V –p (b) Va − Vb = + ve . +r Hence, Va > Vb q2 – 6 V (c) Current flows through two resistors, i = 18 − 0 = 2A 6+3 ∴ Vb − 0 = iR = 2 × 3 q1 = 72 µC or Vb = 6V q2 = 18 µC (d) Initially, V3µF = V6µF = 18V Charge flow from S = (Final charge on plates p ∴ q3µF = 54 µC (q = CV ) and r) − (Initial charges on plates p and r) and q6µF = 108 µC Finally, = (− 72 + 18) − (− 36 + 36) = − 54 µC V6µF = V6Ω = iR 37. (a) In steady state, = 2 × 6 = 12 V VC = V ∴ q6µF = 72 µC 2 V3µF = V3Ω = 6V ∴ Steady state charge, ∴ q3µF = 18 µC ∆q = qf − qi = − 36 µC on both capacitors. q0 = CVC = CV 2 36. (a) In resistors (in series) potential drops in direct For equivalent value of τC : We short circuit ratio of resistance and in capacitors (in series) the battery and find the value of Rnet across capacitors and then potential drops in inverse ratio of capacitance. 18 − Va = 6 6 (18) + 3 RR ∴ Va = 6 V R 18 − Vb = 3 (18) 6 + 3 Vb = 12 V Rnet = 3R 2 (c) V3 µF = V3Ω = iR = 18 − 0 (3) = 6V 6+ 3 ∴ τC = CRnet = 3RC ∴ Vb − 0 = 6 V 2 ∴ Vb = 6 V Now, q = q0 (1 − e− t/ τ C )
670 Electricity and Magnetism (b) At t = 0, capacitor offers zero resistance. 3. At t = 0 when capacitors are initially uncharged, Rnet = 3R their equivalent resistance is zero. Hence, whole 2 current passes through these capacitors. ∴ i = V = 2V 4. Changing current is given by 3R/2 3R i = i0 e−t/ τ C iC = iAB = 1 = V or i = V e− t/CR 2 3R R At t = ∞, capacitor offers infinite resistance. So, If we have take log on both sides, we have iC = 0. ibattery = iAB = V ln (i) = ln VR − C1R t ∴ 2R Now, current through AB increases exponentially Hence, ln (i) versus t graph is a straight line with from V to V with same time constant. slope − C1R and intercept + ln VR . 3R 2R Intercepts are same, but |slope|1 > |slope|2. (i- t) graph is as shown below. 5. During charging of a capacitor 50% of the energy i supplied by the battery is lost and only 50% is stored. V V 2R ∴ Total energy lost = 1 q2 = 1 (EC /2)2 = E2C 6R V 2C 2 C 8 3R t Now, this total loss is in direct ratio r : 2r or 1: 2 (i - t) equation corresponding to this graph is ∴ Energy lost in battery is 1 rd of E2C . i = V + V (1 − e− t/ τ C ) 38 3R 6R 6. Equal and opposite charges should transfer from LEVEL 2 two terminals of a battery. For charging of a Single Correct Option capacitor, it should lie on a closed loop. +σ +3σ ER 1. 7. C E1 E2 E3 –ve Ea b +ve i σ 3σ 2σ E0 R0 2ε0 2ε0 ε0 E1 = + = = E3 E2 = 3σ − σ = σ i = [E − E0 ] 2ε0 2ε0 ε0 R + R0 E1 and E2 are in the negative direction and E3 in Now, Va − E + E0 + iR0 = Vb positive direction. ∴ Va − Vb = (E − E0) − iR0 2. Let E be the external field (toward right). Then, = (E − E0 ) − R R0 1 + R0 E− σ =8 …(i) 2ε0 = R (E − E) R + R0 E + σ = 12 …(ii) 2ε0 CR (E − E0) ∴ q =C (Va − Vb ) = R + R0 Solving these equations, we get σ = 4ε0
Chapter 25 Capacitors 671 8. Initially = 1 × 1 (100)2 2 10 ln 2 C net = (C0) (C0) = C0 C0 + C0 2 = 500 J ln 2 = 0.5 C0 Finally 12. Net capacitance between points A and P will be C net = (C0 /2) (2C0) equal to the net capacitance between points P (C0 /2) + 2C0 and B. = 0.4 C0 13. Total charge = (2C ) (4V ) − CV 9. The simple circuit is as shown below. = 7CV R C Common potential after they are connected is R VC = Total charge Total capacitance VR = 7 CV = 7V 2C + C 3 Rnet = R Heat = U i − U f 3 = 1CV 2 + 1 (2C ) (4V )2 22 ∴ τC = CRnet = CR 3 1 7 2 2 3 q = q0 (1 − e− t/ τ C ) , − × 3C × V where, q0 = CV = 25CV 2 3 10. Common potential in parallel grouping, 14. Total heat produced = 1CV 2 V = Total charge Total capacity 2 = 1 (2 µF) (5)2 = (2 × 100) + (4 × 50) 2+ 4 2 = 200 V = 25 µJ 3 Now, this should distribute in inverse ratio of Loss = U i − U f resistors, as they are in parallel. = 10− 6 1 × 2 × 100 × 100 + 1 × 4 × 50 × 50 ∴ H 5Ω = R 2 2 HR 5 1 200 × 2300 or H 5Ω = R (Total heat) 2 3 R + 5 − × 6 × = 1.7 × 10− 3J or 10 = R R (25) + 5 11. V0 = i0R = (10) (10) = 100 V Solving this equation, we get R = 130 Ω After 2 s, current becomes 1 th. Therefore, after 4 15. In position-1, initial maximum current is 1 s, current will remain half also called half-life. t1/ 2 = (ln 2) τC = (ln 2) CR i0 = V = 10 = 2A ∴ C = (t1/ 2) = 1 F R 5 (ln 2)R 10 ln 2 At the given time, given current is 1A or half of the above value. Hence, at this is instant capacitor is Total heat = 1 CV02 also charged to half of the final value of 5 V. 2
672 Electricity and Magnetism Now, it is shifted to position-2 wherein steady 22. V1 = 1.5 state it is again charged to 5V but with opposite V1.5 1 polarity. 1.5 1.5 + 1 Ui =U f = 1CV 2 (Q V = 5 V ) ∴ V1 = (30) 2 ∴ Total energy supplied by the lower battery is = 18 V converted into heat. But double charge transfer V2.5 = 0.5 = 1 V0.5 2.5 5 (from the normal) takes place from this battery. 1 1 + 5 ∴ Heat produced = Energy supplied by the ∴ V2.5 = (30) battery = (∆q)V = (2CV )(V ) = 2CV 2 =5V Now, |Vab | = V1 − V2.5 = 2 × 2 × 10− 6 × (5)2 = 13 V = 100 × 10− 6J 23. In the figure, = 100 µJ 16. Equivalent capacitance of 6 µF and 3 µF is also +6 V 2 µF and charge across it is also q or circuit is – balanced. Hence, there is no flow of charge. + 17. Two capacitors are in parallel. 60 µF q1 ∴ U = 1 C V 2 2 net = 1 (2C )V 2 = CV 2 – q2 V (let) 2 + 20 µF 30 µF = ε 0A V 2 +q3– d +2 V +3 V 18. Initially, the rate of charging is fast. q1 + q2 + q3 = 0 19. V1Ω = 5 + 2 = 7 V 60 (V − 6) + 20 (V − 2) + 30 (V − 3) = 0 Solving this equation, we get ∴ i1Ω = V = 7A R V = 49 V 11 V2µF = 6 V ∴ q2µF = CV = 12 µC 20. During charging capacitor and resistance of its A 3 µF 1 µF C 3 µF B wire are independently connected with the battery. 24. 2 µF Hence, τC = CR 1 µF During discharging capacitor is discharged through both resistors (in series). Hence, τC = C (2R) = 2CR 10 V 21. Total charge = 3 × 100 − 1 × 100 = 200 µC Common potential (in parallel) after S is closed, is VAB = (C )BC = 3 = 1 V = Total charge VBC (C )AB 6 2 Total capacity = 200 µC ∴ VAB = 1 (10) V 4 µF 1 + 2 = 50 V = 10 V 3
Chapter 25 Capacitors 673 25. Applying Kirchhoff’s loop law in outermost loop, = Kε0 σ we have 3 µF 5 × 8.86 × 10− 12 +– 7.4 × 10− 12 = q =6s 18 V 4 Ω 15 V 1Ω 5Ω 2r 32. The given time is the half-life time of the exponentially decreasing equation. q ∴ t = t1/ 2 = (ln 2) τC = (ln 2) CRnet –+ 3 A 2 µF 2.5 A ∴ Rnet = (ln t 2) C − q + 15 − 2 × 2.5 − q − 3 × 1 + 18 = 0 = 2 (ln 2) µs = 4Ω 32 (ln 2) (0.5 µF) Solving this equation, we get ∴ Resistance of ammeter = 2Ω q = 30 µC 33. Four capacitors are in parallel charge across each 26. τC = CR = 6 s is q = CV . Two surfaces of plate C marks two capacitors, one with B and other with D and C is Now, q0 = CV = 10 µC connected to positive terminal of the battery. q = q0e− t/ τ C = (10 µC) e− 12/6 Hence, = 1e 2 (10 µC) qC = 2CV = + 40 µC 12 34 = (0.37)2 (10 µC) 34. 27. q = (E1 + E2) Cnet = (E1 + E2 ) C 1C 2 C2 + C2 = q = E1 + E2 q1 = q4 = qtotal = CV − CV + Q = Q C2 C1 + C2 2 2 2 Vab C1 28. H 1 = H 2 = U i − U f q2 = (Q + CV ) − Q = Q + CV 2 2 The only change is by increasing the resistance τC q3 =− q2 = − Q + CV increase. Hence, process of redistribution of 2 charge slows down. 29. Just after the switch is closed C1 is short-circuited Electric field between two plates and hence the and current passes through R1 and C1 only. potential difference is due to q2 and q3 only. PD = q2 = V + Q 30. i1 = 2VR −t C 2C e 6 CR −t More than One Correct Options ∴ i2 = V e CR 1 23 4 R Q –Q 1. Q 22 Q 5t 2 2 ∴ i1 = e 6 CR i2 2 We can see that this ratio is increasing with time. qtotal =Q 2 2 31. τC = CR q1 = q4 = = K ε0 A Adσ R = σlA q2 = Q − q1 = Q d 2
674 Electricity and Magnetism q3 = − q2 = −Q Hence, heat is produced at the same rate in A 2 and B. EA = E1 + E4 (towards left) Further, in steady state =Q + Q =Q 4 Aε0 4 Aε0 2Aε0 VC = VB = ε 2 EC = − EA ∴ U = 1 CVC2 = 1 Cε2 =Q 2 8 2 Aε 0 (towards right) σ = q2 2ε0 2ε 0 A 5. F = qE = (q) EB = E2 + E3 (towards right) =Q + Q=Q 4 Aε0 4 Aε0 2Aε0 q remains unchanged. Hence, F remains unchanged. 2. In steady state, E= σ = q qC = EC and q2C = 2EC ε0 Aε0 τC = 2CR of both circuits At time t , q remains unchanged. Hence, E also remains unchanged. qC = EC (1 − et/ τ C ) U = q2 or U ∝ 1 q2C = 2EC (1 − e− t/ τ C ) 2C C C will decrease. Hence, U will increase. ∴ qC = 1 V = Ed or V ∝ d q2C 2 d is increasing. Hence, V will increase. 3. In steady state, current through capacitor wire is 6. Ci = (C ) (2C ) = 2C C + 2C 3 zero. Current flows through 200 Ω, 900 Ω and A2. VC = q = 4 × 10− 3 qi = CiE = 2 EC C 100 × 10− 6 3 = 40 V C f = 2C ∴ qf = 2EC This is also potential drop across 900 Ω resistance and 100 Ω ammeter A2 (Total resistance ∆q = qf − qi = 1000 Ω). Now, this 1000 Ω and 200 Ω are in = 4 CE series. Therefore, 3 V2 = V200 Ω = V1000 Ω 7. Let (+ q) µC charge flows in the closed loop in 5 clockwise direction. Then, final charges on = 40 = 8 V different capacitors are as shown in figure. 5 +– Emf = V1000 Ω + V200 Ω = 48 V qq + i = Emf +– +– Net resistance (300 – q ) (360 – q) = 48 = 1 A Now, applying Kirchhoff’s loop law 1200 25 360 − q + 300 − q = q 3 2 1.5 4. Current through A is the main current passing Solving the above equation, we get through the battery. So, this current is more than q = 180 µC the current passing through B. Hence, during charging more heat is produced in A. 8. If the battery is disconnected, then q = constant In steady state, C = ε0A or C ∝ 1 dd iC = 0 and iA = iB
Chapter 25 Capacitors 675 d is decreased. Hence, C will increase. Hence, charge on C2 at any time t is U = 1 q2 or U ∝ 1 2C C qC2 = C 2 q0 (1 − e−t/ τ C ) C1 +C C is increasing. Hence, U will decrease. 2 V = q or V ∞ 1 CC Initially, C2 is uncharged so, whatever is the charge on C2, it is charge flown through switches. C is increasing. Hence, V will decrease. 2. Common potential in steady state when they 9. (a) At t = 0, emf of the circuit = PD across the finally come in parallel is capacitor = 6 V. ∴ i = 6 = 2A V = Total charge = q0 Total capacity C1 + C2 1+ 2 Total heat dissipated = U i − U f q02 1 q0 2 2C 1 2 C1 + Half-life of the circuit = − (C 1 + C2) C = (ln 2) τC (ln 2) CR = (6 ln 2) s. 2 In half-life time, all values get halved. = q02 C 1C 2 C1 +C 2 For example 2C1 VC = 6 = 3V 3. Eair = E0 = V 2 d i = 2 = 1A 4. Edielectric = E0 = V 1 K Kd ∴ V1Ω = iR = 1V Match the Columns V2Ω = iR = 2V 4 × 4 1µF 4 µF 9 µF 1. (a) Ci = 4 + 4 = 2 µF 10. V1 V2 V3 Cf = C 1C 2 = 8× 2 C1 + C2 8+ 2 V∝1 = 1.6 µF C In series, (as q = constant) q = CV ∴ or V2 = 1 Since, total capacity is decreasing. Hence, V1 4 charge on both capacitors will decrease. ∴ Now, V1 10 (b) U2 = 1 q2 4 4 2 C V2 = = = 2.5 V V3 = 1 q has become 1.6 or 0.8 times but C is halved. V1 9 2 Hence,U 2 will increase. V3 = V1 = 10 V (c) V2 =q 9 9 C E = 10 + 2.5 + 190 V q has become 0.8 times and C is halved. Hence, V2 will increase. Comprehension Based Questions (d) E2 = V2 or E2 ∝ V2 d 1. Finally, the capacitors are in parallel and total 2. (a) Ci = 2 µF charge (= q0) distributes between them in direct ∴ qi = 60 µC ratio of capacity. C f = 6 µF ∴ qC2 = C2 q0 → in steady state. ∴ qf = 180 µC C1 + C 2 ∴ ∆q from the battery = qf − qi =120 µC. But this charge increases exponentially.
676 Electricity and Magnetism (b) Between 4 µF and 2 µF charge distributes 5. Let C0 = ε0A indirect ratio of capacity. Hence, on 2 µF d qi = 2 (60 µC) = 20 µC C1 = ε 0 ( A / 2) + K ε 0 ( A / 2) 2 + 4 d d qf = 2 (180) 60 µC = C0 + C0 = 3C 0 + 4 2 2 2 ε0 A 2 ε0 A ∴ ∆q = ∆f − qi = 40 µC C2 = d/2 + = 1 + (c) On 3 µF, initial charge is 60 µF and final charge d/2 1 d − K d K is zero. 4 ε0A 4 ∴ ∆q = 60 µC = 3 d = 3 C0 (d) On 4 µF ∴ C1 = 9 C2 8 qi = 2 4 (60 µC) = 40 µC + 4 Capacitors are in series. qf = 2 4 (180 µC) = 120 µC Hence, q1 = q2 + 4 or q1 = 1 q2 ∴ ∆q = qf − qi = 80 µC or ∴ U = 1 q2 3. (a) In second figure, VC1 = V = maximum 2C 6. Hence, qC1 is maximum. U∝1 =V C (b) In first figure, VC2 3 (as q is same) In second figure, VC2 = V U1 = C2 = 8 U2 C1 9 In third figure, VC2 = C V =V 2C + C 3 = qtotal q1 = q8 2 = 7Q In fourth figure, VC2 = C 2C V = 2V q2 = 4Q − q1 = − 3Q + 2C 3 q3 = − q2 = + 3Q q4 = Q − q3 = − 2Q Now, q = CV q5 = − q4 = + 2Q q6 = 2Q − q5 = 0 Hence, qC2 is minimum in first and third q7 = − q6 = 0 figures. (c) In second figure, VC1 = V = maximum (d) Similar to option (b) Subjective Questions 4. After closing the switch, the common potential is 1. V3 5 parallel. V2 – V = Total charge = CV = V 4 Total capacity 3C 3 V1 1C V 2 1 CV 2 3 18 3 UC = = 2 2+ 1 V 2 1 CV V3 1 2 3 9 U 2C = (2C ) = 2 V1 Loss of energy = U i − U f C = ε0A d 1CV 2 1 V 2 2 2 3 (between two successive plates). = − × 3C × The effective capacity has to be found between V1 = 1CV 2 and V2. 3
Chapter 25 Capacitors 677 C CV − q + 4 CV − q + 9CV − q V1 V2 C 2C 3C V0 16CV − q 4C V1 V3 + = 0 C + C V3 V2 – or q = 4.8 CV = 24 CV 5 C V2 V1 V1 V3 2V 0 Now, V1 = V −q =V − 24 V = − 19 V C 5 5 V0 3 3 V2 = 2V − q = 2V − 24 V =− 2V 2C 10 5 C net =C + (2C )(C ) = 5C = 5 ε0A 2C + C 3 3d V3 = 3V − q = 3V − 24 V = 7V and 3C 15 5 q3 = + [C (V1 − V2) + C (V1 − V3)] q − 6V = 14 V = ε0A V0 + V0 = 4 ε 0 AV0 V4 = 4V − 4C = 4V 5 5 d 3 3d 4. At t = 5 ms,V = 10 V ε 0A 2V0 q5 = C (V3 − V2) = d 3 V 10 ∴ iR = R = 4 = 2.5 A Ans. = 2ε0AV0 Further, q = CV = (300 × 10−6)(2000t) = 0.6t 3d iC = dq = 0.6 A = constant Ans. 2. (a) V = qnet = C1V0 = V0 dt Cnet C1 + C2 1 + C2/C1 5. Potential energy stored in the capacitor, = 120 = 80 V U = 1CV 2 = 1 × 5 × 10−6 × (200)2 = 0.1 J 1 + 4/8 22 (b) Ui = 1 C 1V02 = 1 × 8 × 10−6 × (120)2 During discharging this 0.1 J will distribute in 2 2 direct ratio of resistance, = 5.76 × 10–2 J ∴ H 400 = 400 × 0.1 400 + 500 Uf = 1 + C2)V 2 2 (C1 = 44.4 × 10−3 J = 1 × 12 × 10−6 × (80)2 = 44.4 mJ Ans. 2 6. = 3.84 × 10–2 J 6Ω 3Ω 3. Let + q charge rotates in the loop in clockwise E/3 + – direction for achieving equilibrium state. In final steady state, charges on the capacitors will be as shown below –+ 4Ω 4Ω E/2 (CV – q) – E (4CV – q) + (a) Current in lower branch = E/8 = 3 A (16CV – q) + Current in upper branch = E/9 = 24 /9 = 2.67 A – (b) PD across the capacitor = E/2 − E/3 = E/6 From q = CV , we have 16 = (4) E +– 6 (9CV – q) ∴ E = 24 V Now, applying Kirchhoff’s loop law we have
678 Electricity and Magnetism (c) After short-circuiting the battery, we will have 9. (i) Charge on capacitor A, before joining with an to find net resistance across capacitor to uncharged capacitor. calculate equivalent value of τC in discharging. 3Ω and 6Ω are in parallel. Similarly, 4 Ω and 2 µF 4 Ω are in parallel. They are then in series. 34 +– ∴ Rnet = 4Ω, τC = CRnet = (4 × 4) µs = 16 µs q2 During discharging q = q0e−t/ τ C 2+ –5 3 µF 1 – q1 q3 2 µF or 8 = 16e−t/ 16 +6 Solving this equation, we get t = 11.1µs Ans. qA = CV = (100)(3) µC = 300 µC Similarly, charge on capacitor B 7. 2C C –+ –+ qB = (180)(2) µC = 360 µC q1 q2 Let q1, q2 and q3 be the charges on the three – capacitors after joining them as shown in figure. C (q1, q2 and q3 are in microcoulombs) + q1 – q2 From conservation of charge Net charge on plates 2 and 3 before joining = net charge after joining 110 V 110 V ∴ 300 = q1 + q2 …(i) Applying loop law in two closed loops, we have Similarly, net charge on plates 4 and 5 before 110 − q2 + q1 − q2 = 0 joining = net charge after joining CC or q2 = (110C ) − 360 = − q2 − q3 or 360 = q2 + q3 and − 110 + q1 + q1 − q2 = 0 …(ii) 2C C Applying Kirchhoff’s second law in closed loop or q1 = 440C q1 − q2 + q3 = 0 3 32 2 or 2q1 − 3q2 + 3q3 = 0 …(iii) Solving Eqs. (i), (ii) and (iii), we get Potential difference between points M and N is VN − VM = q1 − q2 q1 = 90 µC C q2 = 210 µC = 110 volt 3 Ans. and q3 = 150 µC 8. Let us first find charges on both capacitors before (ii) (a) Electrostatic energy stored before and after closing the switch. completing the circuit, C1 C1 Ui = 1 (3 × 10−6)(100)2 + 1 (2 × 10−6)(180)2 2 2 –+ q1 U = 1 CV 2 q0 q2 2 EE E + 2 C2 – q0 + = 4.74 × 10−2 J or U i = 47.4 mJ C2– 1 (b) Electrostatic energy stored after completing the circuit, q2 = EC2 and q1 = 0 Uf = 1 (90 × 10−6)2 + 1 (210 × 10−6)2 2 (3 × 10−6) 2 (2 × 10−6) q0 = E C 1C2 C1 +C 2 + 1 (150 × 10−6)2 = 1 q2 2 (2 × 10−6) U 2 From 2, −q0 charge will flow, so that charge on C right hand side plate of C1 becomes zero. From 1, q2 charge will flow. = 1.8 × 10−2 J or U f = 18 mJ Ans.
Chapter 25 Capacitors 679 10. (a) In steady state, capacitors will be in parallel. 13. (a) Let q be the charge on smaller sphere. Then, Charge will distribute in direct ratio of their +1 µC capacity. +1 µC –1 µC ∴ q1 = C C1 q0 1+C 2 and q2 = C C2 q0 1+C 2 Initial emf in the circuit is potential difference across capacitor C1 or q0/C1. Therefore, initial current would be i0 = q0 /C1 = q0 Vinner = 0 R C 1R ∴ Kq + K (2) = 0 or q = − 1µC Current as function of time will be i = i0e−t/ τ C 24 Here, τC = C 1C 2 R Ans. Now, Vouter = K (2 − 1) × 10−6 C1 +C 4 × 10−2 2 (b) Ui = 1 q02 and Uf = 1 q02 = 9 × 109 × 10−6 2 C1 2 C1 + C2 4 × 10−2 Heat lost in the resistor = 2.25 × 105 V Ans. =Ui −U f = q02 C2 Ans. (b) Charge distribution is as shown in above 2 C1(C1 + figure. C 2 ) 90 11. τC = CR = Kε0 A ρAd = Kε0ρ 14. (a) Fig. (a) V6 = 3 = 30 V, q6 = 6 × 30 = 180 µC d 90 = 5 × 8.86 × 10−12 ≈ 6 s V3 = 3 = 30 V 7.4 × 10−12 q3 = 30 × 3 = 90 µC Ans. Initial current, Fig. (b) Capacitor 1µF is short-circuited. i0 = q0 / C = q0 = q0 = 8.55 = 1.425 µA Therefore, q1 = 0. R CR τC 6 20 Now, current as function of time i = i0e−t/ τ C V26 = 20 + 20 + 10 × 100 = 40 V or i = (1.425)e−12/6 = 0.193 µA Ans. This 40 V will distribute in inverse ratio of 12. (a) C = ε0A , U = Q 2 = Q 2x capacity. x 2C 2ε0A ∴ V6 = 2 × 40 = 10 V (b) dU = Q 2 8 dx 2ε0A V2 = 6 × 40 = 30 V 8 Q2 ∴ dU = dx ∴ q6 = 60 µC, q2 = 60 µC Ans. 2ε0A (b) Fig. (a) When S is open, 6 µF is short-circuited Q2 (c) dx = dW = Fdx or V6 = 0, q6 = 0 2ε0A and V3 = 90 V, q3 = 270 µC Ans. ∴ F = Q2 Fig. (b) When S is open, V1 = 100 V 2ε 0 A q1 = 100 µC (d) Because E between the plates is due to both V26 = 100 V the plates. 2 While F = (Q ) (field due to other plate) Ans. ∴ V6 = 8 × 100 = 25 V
680 Electricity and Magnetism and V2 = 6 × 100 = 75 V ∴ Rnet = 3+ 4 × 6 = 5.4 MΩ 8 4 + 6 ∴ q6 = 150 µC, q2 = 150 µC ∴ τC = CRnet = (10 × 10−6)(5.4 × 106) Further, VA − 0 = V2 = 54 s ∴ VA = V2 = 75 V Ans. ∴ V2 = 6 + (10.8 − 6)(1 − e−t/C ) 15. Current in the circuit, i = 10 = 1 A = 6 + 4.8(1 − et/ 54) 4 + 1+ 2+ 3 Here, V2 is in kV and t is second. Now, V5 µF = V1,2Ω = 3 V 17. Circuit can be drawn as shown in figure. ∴ q5 = 15 µC Further, R2 C ∴ V3µF = V2, 3Ω = 5 V R3 q3 = 15 µC Ans. E 16. (a) At t = 0, when capacitor is uncharged, its R1 equivalent resistance is zero. ∴ Rnet = 4 + 6× 3 = 6 MΩ In charging of capacitor, R3 has no role. 6+ 3 In steady state, potential difference across or i1 = 18 × 103 A = 3 mA capacitor = potential difference across R2 = E/2 6 × 106 Therefore, steady state charge across capacitor This will distribute in inverse ratio of q0 = CE 2 resistances. ∴ i2 = 3 3 i1 = 1 mA and i3 = 2 mA To find time constant of circuit we will have to short 6+ circuit the battery, then we will find net resistance At t = ∞, when capacitor is completely across capacitor. charged, equivalent resistance of capacitor is Rnet = R ⇒ τC = CRnet = CR 2 2 infinite. ∴i3 = 0, i1 = i2 = 18 × 103 = 1.8 mA Ans. R (4 + 6) × 106 (b) At t = 0, R V2 = i2R2 = (1 × 10−3)(6 × 106) V ∴ Charge in the capacitor at time t would be = 6 kV 2t At t = ∞, q = q0(1 − e− t / τC ) = CE (1 − − CR ) Ans. 2 V2 = i2R2 = (1.8 × 10−3)(6 × 106) V e = 10.8 kV Ans. 18. q2 = 20 µC (c) To find time constant of the circuit we will ∴ q1 = 10 µC (as they are in parallel) have to short-circuit the battery and find resistance across capacitor. In that case, R1 and Energy stored at this instant, R2 are in parallel and they are in series with R3. U = 1 q12 + 1 q22 V2 (kV) 2C1 2C2 10.8 = 1 × (10−5 )2 + 1 × (2 × 10−5)2 2 10−6 2 2 × 10−6 = 1.5 × 10−4 J 6 = 0.15 mJ t In charging of a capacitor 50% of the energy is stored and rest 50% is dissipated in the form of heat.
Chapter 25 Capacitors 681 Therefore, 0.15 mJ will be dissipated in the form 20. When S1 is closed and S2 open, capacitor will of heat across all the resistors. In series in direct discharge. At time t = R1C , one time constant, ratio of resistance (H = i2Rt) and in parallel in charge will remain q1 = 1e times of CV or inverse ratio of resistance. ∴ H 2 = 0.075 mJ, H 3 = 0.05 mJ q1 = CV e and H6 = 0.025 mJ Ans. 19. (a) At t = 0, capacitor is equivalent to a battery of When S1 is open and S2 closed, charge will increase (or may decrease also) from CV to CE emf E. 2 e Net emf of the circuit = E − E/2 = E/2 exponentially. Time constant for this would be Total resistance is R. (R1C + R2C ). Charge as function of time would be q = qi + (qf − qi )(1 − e−t/ τ C ) Therefore, current in the circuit at t = 0 q = CV + CE − CV (1 − e− t / τ C ) would be e e i = E/2 = E Ans. R 2R (b) Let in steady state there is total q charge on C. After total time 2R1C + R2C or t = R1C + R2C , one time constant in above equation, charge will Initial charge on C was CE/2 . Therefore, remain charge on 2C in steady state would be CE − q with polarities as shown. This is q = CV + CE − CV 1 − 1e 2 e e because net charge on lower plate of C and of = EC 1 − 1e + VC e2 upper plate on 2C should remain constant. Applying loop law in the circuit in steady 21. At t = 0, capacitor C0 is like a battery of state, we have emf = Q0 = 1 V C0 E − q + CE/2 − q = 0 C 2C ∴ q = 5CE Net emf of the circuit = 4 − 1 = 3 V 6 Total resistance is R = 100 Ω R ∴ Initial current = 3 = 0.03 A + 100 Cq – This current will decrease exponentially to zero. ∴ i = 0.03 e−t/ τ C E 2C – CE –q Here, τC = Cnet R = (1 × 10−6)(100) + 2 = 10−4 s Therefore, charge on C increases from qi = CE ∴ i = 0.03 e−104 t Ans. 2 22. From O to A to qf = 5CE exponentially. 6 VC Equivalent time constant would be A τC = C × 2C R = 2 CR C + 2C 3 Therefore, charge as function of time would be O t q = qi + (qf − qi )(1 − e−t/ τ C ) VC = at (a = constant) = CE + CE − 3t Ans. ∴ qC = CVC = Cat 2 3 1 − 2C R e
682 Electricity and Magnetism ∴ i = dqC = aC From A onwards When V = constant (say V0) dt V0 = at or t = V0 VR = iR = aCR = constant a From A onwards VC = constant ∴ VED = aCR (1 – e–V0 / aCR ) ∴ qC = constant ∴ i = dqC = 0 After this VED will decrease exponentially. Hence, a rough graph is as shown in figure. dt or VR = 0 VED Therefore, VR versus t graph is as shown in figure. VR t Ot 24. qi = CVi = 100 µC, qf = CV f = − 50 µC 23. From O to A V = at Therefore, charge will vary from 100 µC to − 50 µC exponentially. Here, a is a positive constant. ∴ q = − 50 + 150 e−t/ τ C , Here q is in µC ∴ at = q + iR τC = CR = (10−6)(5 × 103) = 5 × 10−3 C ∴ q = − 50 + 150 e−t/ τ C Differentiating w.r.t. time, we have VC = q = (− 50 + 150e−t/ τ C )V C a= 1 ddqt + ddti R C or VC = 50 (3e−200t − 1) or ddit R =a– i as i = ddqt i = − dq = 150 × 10−6 e−200t C dt τC ∫ ∫∴ i di = t dt = 150 × 10−6 e−200 t = 30 × 10−3e−200t 0 a – i/C 0 R 5 × 10−3 ∴ i = aC (1 – e–t/CR ) VR = iR = 150e−200t Ans. V 25. At t = 0, equivalent resistance of an uncharged AB capacitor is zero and a charged capacitor is like a battery of emf = potential difference across the capacitor. (a) VC1 = q1/C1 = 2 V t ∴ Net emf of the circuit = 9 − 2 = 7 V O or 9 + 2 = 11 V i.e. current in the circuit increases exponentially Net resistance = 30 + 30 × 60 = 50 Ω ∴ VED = iR = aCR (1 – e–t/CR ) or VED also increases exponentially. 30 + 60 q ∴ Current at t = 0 would be i0 = 7 A + + –i 50 or 11 A Ans. 50 V = at R (b) In steady state, no current will flow through – the circuit. C2 will therefore be short-circuited, while PD across C1 will be 9 V. ∴ Q2 = 0 and Q1 = 9 µC Ans.
Chapter 25 Capacitors 683 26. Time constant of the circuit is For t ≥ 250 µs i = – i0′ e–t/ τ C = – 0.11 e–4000 t τC = CR = (0.5 × 10–6) (500) = 2.5 × 10–4 s For t ≤ 250 µs i = i0e–t/ τ C The (i - t) graph is as shown in figure. Here, i0 = 20 = 0.04 A i (A) 500 0.04 ∴ i = (0.04 e–4000t ) amp At t = 250 µs = 2.5 × 10–4 s 0.015 –0.11 i = 0.04 e–1 = 0.015 amp 2.5 t (× 10–4 s) At this moment PD across the capacitor, 27. Capacitor C1 will discharge according to the VC = 20 (1 – e–1) = 12.64 V equation, So when the switch is shifted to position 2, the current in the circuit is 0.015 A (clockwise) and PD across capacitor is 12.64 V i = 0.015 A q = q0e−t/ τ C …(i) Here, τC = C1R 500 Ω and discharging current : 20 V 40 V + 12.64 V i = − dq = q0 .e−t/ τ C = q0e−t/ τ C …(ii) – dt τC C 1R At the given instant i = i0 Therefore, from Eq. (ii) q0e−t/ τ C = i0C1 R at this instant or charge C1 at this instant will be 500 Ω q = (i0C1R) [From Eq. (ii)] ⇒ 40 V + Now, this charge q will later on distribute in C1 12.64 V and C2. – q2 ∴ Ui = 2C 1 As soon as the switch is shifted to position 2 and q2 C2) = U f 2(C1 + current will reverse its direction with maximum ∴ Heat generated in resistance, current. i0′ = 40 + 12.64 H =Ui −Uf 500 Substituting values of q :U i andU f , we get = 0.11 A H = (I0R)2C1C2 Ans. 2(C1 + C2) Now, it will decrease exponentially to zero.
26. Magnetics INTRODUCTORY EXERCISE 26.1 4. Magnetic force may be non-zero. Hence, 1. Q qE = Bqv sin θ acceleration due to magnetic force may be [E/ B ] = [v ] = [LT−1 ] non-zero. ∴ Magnetic force is always perpendicular to 2. From the property of cross product, F is always velocity. Hence, its power is always zero or work done by magnetic force is always zero. Hence, it perpendicular to both v and B . can be change the speed of charged particle. 3. May be possible that θ = 0° or 180° between v and 5. F = q (v × B) B, so that Fm = 0 F is along position y -direction. q is negative and v is along positive x -direction. Therefore, B should 4. F = q (v × B) be along positive z -direction. Here, q has to be substituted with sign. 6. (a) r = mv 5. Apply Fleming’s left hand rule. Bq 6. Q F = Bqv sinθ or r ∝ m as other factors are same. (b) f = Bq ∴ v= F Bqsin θ 2πm or f ∝ 1 = 3.5 × 10−3 4.6 × 10−15 × sin 60° × 1.6 × 10−19 m = 9.47 × 106 m/s 7. Q r = 2qVm Bq 7. Q F = Bqv sin 90° = 0.8 × 2 × 1.6 × 10−19 × 105 = 2.56 × 10−14 N INTRODUCTORY EXERCISE 26.3 1. Q l = l $i INTRODUCTORY EXERCISE 26.2 Now, F = i(l × B) = i [(li$) × (B0$j + B0k$ )] 1. Path C is undeviated. Therefore, it is of neutron’s = ilB0(i$ − $j) = | F | = 2ilB0 path. From Fleming’s left hand rule magnetic force 2. No, it will not change, as the new $i component of on positive charge will be leftwards and on negative charge is rightwards. Therefore, track D B is in the direction of l. is of electron. Among A and B one is of proton and other of α-particle. 3. Q i = 3.5 A Further, r = mv or r ∝ m l = (− 10−2$j) Bq q Now, apply F = i (l × B) in all parts. m m q α q 4. C Since, > P ∴ rα > rP or track B is of α-particle. 2. r = 2km or r ∝ m ( k,q and B are same) AD Bq FACD = FAD ∴ |Fnet | = 2|FAD | = 2ilB mp > me ⇒ rp > re = 2 × 2 × 4 × 2 = 32 N 3. The path will be a helix. Path is circle when it enters normal to the magnetic field.
Chapter 26 Magnetics 685 INTRODUCTORY EXERCISE 26.4 (b) 2πR = 4(0.4) 2R = 1.6 m 1. Q M= q π L 2m B = µ0i = (4π × 10−7 ) (10) ∴ 2R (1.6 /π ) M = 2qm L = 2qm (Iω) = 24.7 × 10−6 T = 2qm 1 mR 2 (ω) = qR2ω 2. Magnetic field due to horizontal wire is zero. 2 4 Magnetic field due to vertical is 2. Q M = iA = i(πR2) B = µ0 i (sin 90° + sin 0° ) 4π x = (0.2) (π ) (8 × 10−2)2 = µ0 i = (4.0 × 10−3) A -m2 4π x (inwards) Now, M = MM$ = (4.0 × 10−3) (0.6i$ − 0.8 $j) 3. Both straight and circular wires will produce (a) τ = M × B magnetic fields inwards. (b) U = − M ⋅ B ∴ B = µ0 i + µ0i 2π R 2R 3. Q L = 2πR ⇒ R = L 2π = (2 × 10−7 ) (7) + (4π × 10−7 )(7) 0.1 2 × 0.1 M = iA = i (πR2) L 2 iL2 = 5.8 × 10−5 T 2π 4π = (i) (π ) = 4. Magnetic field at O due to two straight wires = 0 τ max = MB sin 90° = iBL2 Magnetic field due to circular wire, 4π B = 1 (due to whole circle) 4. Q ∆U = U 0° − U 180° 4 = − MB cos 0° + MB cos180° = − 2MB = 1 µ2R0i (inwards) 4 INTRODUCTORY EXERCISE 26.5 = 1 × (4π × 10−7 ) (5) 4 2 × 0.03 1. (a) From screw law, we can see that direction of = 2.62 × 10−5 T magnetic field at centre of the square is inwards 5. Magnetic field at P due to straight wires = 0 as the current is clockwise. Due to circular wires one is outwards (of radius a) and other is inwards. 60° means 1 th of whole 6 circle. ∴ B = 1 µ 0i − µ0i (outwards) 6 2a 2b 45° 45° i =10 A 0.2 m INTRODUCTORY EXERCISE 26.6 1. Applying Ampere’s circuital law, B = 4 µ0 ri (sin α + sin β) BA = µ0 iin 4π 2π r = 4 × 10−7 ×10 (sin 45° + sin 45° ) = (2 × 10−7 ) (1) 0.2 10−3 = 2.83 × 10−5 T = 2 × 10−4 T
686 Electricity and Magnetism This is due to (⋅) current of 1 A. Hence, magnetic 3. Using Ampere’s circuital law over a circular loop lines are circular and anti-clockwise. Hence, of any radius less than the radius of the pipe, we can see that net current inside the loop is zero. magnetic field is upwards. Hence, magnetic field at every point inside the loop will be zero. Bb = µ0 iin 2π r = (2 × 10−7 ) (3 − 1) INTRODUCTORY EXERCISE 26.7 3 × 10−3 = 1.33 × 10−4 T 1. i = NBk A φ This is due to net ⊗ current. Hence, magnetic lines ∴ B = kφ = (10−8 ) (90) are clockwise. NiA 100 × 10−6 × 10−4 2. So, magnetic field at B is downwards. ∫2. B ⋅ d l = µ0(inet ) = 90 T Along path (a), net current enclosed by this path is i = NBk A φ zero. = (0.125 × 10−7 ) (6) (π /180) Hence, line integral = 0 200 × 5 × 10−2 × 5 × 2 × 10−4 Along path (b), inet is ⊗. So, magnetic lines along = 1.3 × 10−7 A this current is clockwise. But, we have to take line integral in counter clockwise direction. Hence, line integral will be negative. Exercises LEVEL 1 6. Q r = 2qVm ⇒ r ∝ m Assertion and Reason Bq q 2. By changing the direction of velocity direction of m and q both are different. Ratio m is not same for q magnetic force will change. So, it is not a constant force. both. 3. To balance the weight, force on upper wire should 7. Q Fe + Fm = 0 qE + q (v × B) = 0 be upwards (repulsion). Further equilibrium can be ∴ checked by displacing the wire from equilibrium ∴ E = − (v × B) = (B × v) position. 8. Q Fm = q(v × B) 4. τ = MB sin 90° ≠ 0 Fm ⊥ v (always) 5. Magnetic field is outwards and increasing with x. and P = F ⋅ v So, magnetic force will also increase with x. The force on different sections are as shown in figure. ∴ Power of magnetic force is always zero. Fe = qE If Fe is also perpendicular to v, then its power is also zero. 9. x x x Force will act in positive x-direction. But, no q+ torque will act. 10. | v| = 2v0 = speed, which always remains constant.
Chapter 26 Magnetics 687 11. R ∝ v, by increasing the speed two times radius 20. Bx = 1 Bc 8 also becomes two times. Hence, acceleration ∴ (= v2/R) will also become only two times. µ 0NiR2 = 1 µ0Ni 2(R2 + x2)3/ 2 8 2R Objective Questions 2. T = 2πm, independent of v. 21. Electric field (acting along $j direction) will change Bq the velocity component which is parallel to B (which is also along $j direction). 3. B = µ0 i, independent of diameter of wire. B = B0$j and v = v0$j will rotate the particle in a circle. Hence, the net path is helical with variable 2π r pitch. 8. In uniform B is force on any current carrying loop 22. r = 2Km ⇒ r ∝ K is always zero. Bq B 9. M = NiA 10. B ⋅ F = 0 as B ⊥ F BC ∴ B⋅α = 0 or 2x + 3 − 4 = 0 23. ∴ x = 0.5 BB + BD BA + BC 12. In uniform magnetic field, force on any current carrying loop is always zero. Bnet 13. r = P or r ∝ 1 (as P = constant) AD Bq q 2qVm 2Vm B1 14. θi = 180° , θ f = 180° − θ 24. r = = Bq q W = Uf −Ui = − MB cos (180° − θ) − (− MB cos 180° ) 26. In (c), two wires are producing u magnetic field = MB cos θ − MB and two wires are producing ⊗ magnetic field. P 27. I1 produces circular magnetic lines current I2 is 37° 37° each small circular element is parallel to (θ = 0°) magnetic field. Hence, force is zero. 15. 5 cm r = 5 cm 28. Arc of radius a 1 th of circle produces magnetic 4 4 cm 3 cm 3 cm field in k$ direction or outwards, while arc of radius b produces magnetic field in − k$ direction. B = µ0 i (sin 37° + sin 37° ) ∴ B= 1 µ 0I k$ + 1 µ 0I (− k$ ) 4π r 4 2a 4 2a 16. Bx = µ 0NiR2 and Bc = µ 0Ni c→ centre. = µ0I 1 − b1 k$ 2(R2 + x2) 2R 8 a 17. F = I (l × B) = I (ba × B) 29. Equivalent current, We can see that all (a), (b) and (c) options are i = q f = ef same. B = µ0i = µ0ef 18. F = Bqv or F ∝ v 2R 2R Now, v = 2qV 30. m ∴ F∝ V C 45° 45° 19. Two fields are additive. a 2 ∴ Bnet = µ0I = 2µ 0I 2 2π (R/2) πR
688 Electricity and Magnetism BC = 4 µ0 I (sin 45° + sin 45° ∴ (7.8 × 10−6) (3.8 × 10−3) (Bx ) 4π a/2 ) = (− 5.2 × 10−3) = 2 2µ0I ∴ Bx = − 0.175 T πa Similarly, 31. At distance r from centre, (7.8 × 10−6) (− 3.8 × 103) (Bz) = 7.6 × 10−3 B = µ 0 (iin ) (From Ampere’s circuital law) 2π r or Bz = − 0.256 T For path-1, iin ≠ 0 (c) From the property of cross product. ∴ B1 ≠ 0 F is always perpendicular to B . For path-2, iin = 0 Hence, F⋅B=0 ∴ B2 = 0 4. Apply F = q (v × B) 32. dB = µ0 i (d l × r) For example, let us apply for charged particle at e. 4π r3 Fe = q v $j − v k$ ( Bi$ ) ∴ dB is in the direction d l × r. Hence, d l is 2 2 × outwards and r is from d l towards P. = qvB (− $j − k$ ) 2 33. Magnetic field due the straight portions is zero. It is only due to arc of circle. ∴ B = φ µ0I (Radius = x) 5. r = 2qVm 2π 2x Bq = µ0I φ ∴ B = 1 2Vm 4πx rq 34. From centre, r = (R − x) =1 2× 2 × 103 × 9.1 × 10−31 0.18 1.6 × 10−19 B= µ0 I r = µ0 I (R − x) 2π R2 2π R2 = 8.38 × 10−4 T 35. From Fleming’s left hand rule, we can see that 6. (a) r = mv ⇒ v = Bqr magnetic force is outwards on the loop. Bq m (b) t = T = πm 2 Bq F (c) r = 2qVm ∴ Bq V = r2B2q2 = r2B2q 2qm 2m So, it tends to expand. 7. (a) Conservation of charge Subjective Questions B 1. F = q(v × B), where q = − 1.6 × 10− 19 C for an They will collide after time, electron and q = + 1.6 × 10− 19 C for a proton t = T = πm 2 Bq 2. F = q(v × B) 3. (a) F = q (v × B) 8. (a) At A, magnetic force should be towards right. or [(7.6 × 10−3) $i − (5.2 × 10−3)k$ ] From Fleming’s left hand rule, magnetic field = (7.8 × 10−6) [(−3.8 × 103)$j (Bxi$ + By$j + Bzk$ )] should be inwards.
Chapter 26 Magnetics 689 Further, r = mv 13. W = Fm or mg = ilB sin 90° Bq ∴ i = mg = (13 × 10−3) (10) ∴ B = mv (r = 5 cm) Bl 0.44 × 0.62 qr = 0.47 A (b) tAB = T = πm 2 Bq Magnetic force should be upwards to balance the weight. Hence, from Fleming’s left hand rule we 9. Component of velocity parallel to B, i.e. vx will can see that direction of current should be from left to right. remain unchanged vy$j and B$j will rotate the 14. (a) ilB = mg particle in y z-plane (⊥ to B). Z or V lB = mg or V = mgR R lB F vy Y = (0.75) (9.8) (25) t 0.5 × 0.45 θ ≈ 817 V vy (b) a = ilB − mg = VlB − g as i = V R At the beginning direction to magnetic force is m mR − k$ [from the relation, F = q (vy$j × Bi$) = (817) (0.5) (0.45) − 9.8 (0.75) (2.0) In the time t, particle rotates an angle ≈ 112.8 m/s2 θ = ω t = Bmq t from its original path. 15. i = 5 A ⇒ B = (0.02$j) T In the figure, we can set that, y - component of Now, applying F = i (l × B) in all parts. Let us velocity at time t is vy cos θ and z - component is find l for anyone parts. − vy sin θ. l cd = rd − rc 10. Fe + Fm = 0 = (0.4 $j + 0.4k$ ) − (0.4 $i + 0.4k$ ) or qE + q (v × B) = 0 = (0.4 $j − 0.4 $i ) or E = − (v × B) 16. Let surface charge density is σ. or E = (B × v) 11. Work is done only by electrostatic force. Hence, f from work-energy theorem r = 1 mv2 = work done by electrostatic force only dq 2 dr = (qE0)z or Speed v = 2qE0 z m Particle rotates in a plane perpendicular to B, i.e. in dq = [(2πr)dr ]σ Equivalent current, xz-plane only. Hence, vy = 0 i = (dq) f 12. When they are moving rectilinearly, net force is dM = iA = [(dq) f ][πr2 ] zero. R ∴ qE = Bqv sin (90° − θ) ∴ M = ∫0 dM ∴ v= E dB = µ0i = µ0(dq) f B cosθ 2r 2r When electric field is switched off, R p= 2Bπqm v cos (90° − θ) ∴ B = ∫0 dB Now, we can find the ratio M . = 2πm E tanθ B qB2
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