Title 17. Electrostatics Easy 2 Ans.pdf ✅

1. (c) C = 4 0R , 4 0 C R =  R m 9 9 = (1 / 9) 9 10 = 10 2. (a) 3. (b) C = 4 0R m C R 9 12 3 0 9 10 10 9 10 4 − − = =   =   Diameter = 2R = 2  9  10–3 = 18  10–3 m 4. (b) C Q V = = KA Qd 0   V  d 5. (a) U CV J 2 1 2 2 8 10 10 (50) 1.25 10 2 1 2 1 − − = =    =  6. (c) FA = FB ; because an uniform electric field produced between the plates. 7. (b) 8 4 10 3.5 10 4 10 ' 3 3 3 =  −   = − = − − − t d t K 8. (a) k E E air medium = 9. (a) Maximum potential difference mm k V V mm k V = 19  0.01 = 0.19 = 190 10. (b) C n C 1 / 3  = (64 ) C 4C 1 / 3 = = 11. (d) 2 12 2 700 10 (50) 2 1 2 1 − U = CV =   J 7 8.7 10 − =  12. (d) ( ) 2 2 1 2 2 1 C C V U = U − U = − J 6 2 2 (10 2) 10 4 10 2 (100 ) − − = −  =  13. (a) 2 12 2 12 10 (50) 2 1 2 1 = =    − U CV J 8 1.5 10 − =  14. (c)   d C 1 1 2 2 1 d d C C =  6 15 2 2 = C  C2 = 45F 15. (d) 16. (c) d A C 0  = and       − + = K t d t A C 0 '   d K t d t C C       − + = ' 3 3 3 3 2 10 2 1 10 2 10 1 10 ' 20 − − − −            −  +  = C  C' = 26 .6F 17. (a) d KA C 0  =  1 2 2 1 2 1 d d K K C C =  (0.4) (0.4 / 2) 2.8 2 1 2 =  C  C2 =11.2  F 18. (c) ( ) ( ) 2 1 1 2 2 1 2 2 1 C C C C V V U + −  = 6 12 2 (3 5) 10 (3 5) 10 (500 300 ) − − +     − = 0.0375 J 8 10 15 10 4 10 6 1 2 4 =     = − − 19. (b) Charge on smaller sphere = Total charge C r r r 10 5 10 5 30 1 2 1  =      + =         + 20. (d) d A C 0  = and d A d d d A C 0 0 2 ( / 2) 2 '   =        − + =  C' = 2C 21. (b) By inserting the dielectric slab. Capacitance (i.e. ability to hold the charge) increases. In the presence of battery more charge is supplied from battery. 22. (a) Initial energy of body of capacitance 4 F is U J i (4 10 )(80) 0.0128 2 1 6 2 =   = − Final potential on this body after connection is 50 . 4 6 4 80 6 30 V = V +  +  = So final energy on it U J f 4 10 (50) 0.005 2 1 6 2 =   = − Energy lost by this body = Ui – Uf = 7.8 mJ 23. (d) 24. (d) Capacitance of the given assembly         − = 2 1 1 2 4 0 R R R R C   ( ) 2 1 1 2 R R R R C −  25. (d) 26. (c) 1 2 2 1 1 d d C C d C   = so C F C 20 4 8 10 2 2 =  = 27. (a) 2 2 1 U = CV so 2 (1200 ) 2 1 24  60  60 = C  C=120 mF
28. (a) Energy density 2 0 2 0 2 2 0 0 2 0 2 2 2 1 2 1 A q E        = =         = = 29. (b) J C Q U 5 5 10 6 2 6 2 8 10 2 10 16 10 2 10 10 (40 10 ) 2 − − − − − =    =    = = 8 10 10 800 erg 5 7 =   = − 30. (c) 31. (b) N Q E d CV F 0.05 2 10 10 2 2 2 6 5 =  =  = = − 32. (d) Work done W = U f − Ui 2 f 0 2 0 .(3 ) 3 ( ) 2 1 and U 2 1 V C Ui = CV = 2 0 2 1 = 3  CV So d AV W 2 0 0  = 33. (b) In the presence of battery potential difference remains constant. Also , d V E = so E remains same. 34. (c) Capacitance with dielectric d K A Cmedium 0  =  d K Cmedium  35. (a) 36. (a) Thin metal plates doesn't affect the capacitance. 37. (b) 2 2 ( ) 2 1 2 1 U = CeqV = nC V 38. (c) UBig n usmall 5 / 3 = 39. (c) After redistribution new charges on spheres are Q C 3 10 10 10 20 10 ' 1   =      + = and Q C 3 20 10 10 20 20 ' 2   =      + = Ratio of charge densities 2 1 2 2 ' 2 ' 1 2 1 r r Q Q =    1 2 10 20 20 / 3 10 / 3 2  =      =        = 2 4 r Q   40. (d) 2 1 / 3 2 2 2 ( ) ( ) r n r nq q r R Q q Big small =  =    4 1 (64) 1 / 3 1 / 3 = = = − − n 41. (a) C R F 1 0 9 0 1 1.1 10 9 10 1 4 −  =   =   = 42. (d) U CV J 2 6 2 4 2 10 (50) 25 10 2 1 2 1 − − = =    =  erg 3 = 25  10 43. (d) U CV 5 10 (20 10 ) 1k J 2 1 2 1 2 6 3 2 = =    = − 44. (d) d A C 0  = As A → 2 1 times and d → 2 times So 4 1 C → times i.e. C C 3 F 4 12 4 1  = = = 45. (c) U CV 6 10 (100 ) 0.03 J 2 1 2 1 2 6 2 = =   = − 46. (c) Because there is no source of charge. 47. (d) d A Cair 0  = , with dielectric slab C=       − + K t d t  0 A Given C = C  3 4 d A K t d t 0 A 0 3  4  =        − + 2 4[( / 2) ] 4( / 2) 4 4 = − = −  = d d d t d t K 48. (d) 2 6 2 10 10 (500 ) 2 1 2 1 = =    − U CV =1.25 J 49. (c) d A C 0  =  A Cd  0 =  0  → m F m Farad m →  2 50. (a) J C Q W 3 2 6 2 1 8 2 32 10 2 100 10 (8 10 ) 2 − − − =     = = 51. (a) V n v (64) 10 160 volt 2 / 3 2 / 3 = =  = 52. (d) Q1 = CV and Q2 = CV Applying charge conservation CV1 +CV2 = Q1 + Q2 CV1 +CV2 = 2CV  V1 + V2 = 2V 53. (a)
54. (c) The given arrangement becomes an arrangement of (n −1) capacitors connected in parallel. So CR = (n −1)C 55. (a) 56. (a) The given circuit is equivalent to a parallel combination two identical capacitors Hence equivalent capacitance between A and B is C = d A d 0A 0   + d 2 0A = 57. (c) 2.4 . 1 2 1 2 F C C C C Ceq =  + = Charge flown = 2.4  500  10–6 C =1200 C. 58. (c) d k A d k A CR C C 1 0 1 2 0 2 1 2   = + = + d A d A 2 4 2 2 0 0   +  = 30F 2 10 4 2 10 = 2 +  = 59. (d) In series combination, charge is same on each capacitor. 60. (b) According to energy conservation, energy remains the same  Uparallel = Useries  2 2 ' 2 1 ( ) 2 1 V n C nC V       =  V' = nV ( V' = potential difference across series combination) 61. (d) The circuit can be drawn as follows CAB F 4 3 3 1 3 1 = +   = 62. (d) In the given system, no current will flow through the branch CD so it can be removed Effective capacitance of the system = 5 + 5 = 10 F 63. (a) 2 1 18 1 9 1 3 1 1 = + + = Cs  Cs = 2F Cp = 3 + 9 +18 = 30 F  15 1 30 2 = = p s C C 64. (b) Total capacitance of given system Ceq F 5 8 = U C V J eq 2 6 6 10 225 180 10 5 8 2 1 2 1 − − = =    =  180 10 10 erg 1800 erg 6 7 =   = − 65. (c) U CV 2 (200 ) 10 0.04 J 2 1 2 1 2 2 6 = =    = − 66. (c) Q1 = Q2 + Q3 because in series combination charge is same on both the condenser and V = V1 + V2 because in parallel combination . V2 = V3 ,Hence V = V1 + V2 67. (b) The given circuit can be drawn as where C = (3 + 2)F = 5F 3 1 60 20 12 1 20 1 5 1 1 = + + = = CPQ  CPQ = 3F 68. (b) In series combination Q is constant, hence according to C Q U 2 2 =  C U 1   1 2 0.3 0.6 1 2 2 1 = = = C C U U 69. (b) Potential difference across 4F capacitor V 500 300 volt 4 6 6  =      + = 70. (c) Charge flowing V C C C C 1 2 1 2 + = . So potential difference across 1 2 1 1 2 1 1 C C C C C V C  + = 1 2 2 C C C V + = 71. (c) In parallel, C = C1 +C2 +C3 = 20F 72. (c) 1 2 3 1 1 1 1 CR C C C = + +  1 1 3 1 2 1 1 ( ) − − − − CR = C +C +C 73. (c) C1 = 2C and / 2, C2 = C so C1 / C2 = 4 : 1 74. (a) In parallel combination V1 = V2 C C A B 1 F 1 F A B 1 F 1 F 3 F A B 10 10 10 10 5 F 5 F C D 12F 20F P Q 5F
or 2 2 1 1 C q C q =  2 1 2 1 C C q q = 75. (c) 76. (d) The circuit can be drawn as follows 77. (a) Energy (U) C q 2 2 = . q remains same so C U 1   = After Before U U 1 1 2 C C + C 78. (a) CAB 4, F 3 3 = 3 + = CAC 3F 2 3 2 3 = + =  CAB : CAC = 4 :3 79. (c) Initial energy 2 2 2 2 1 1 2 1 2 1 Ui = C V + C V , Final energy 2 1 2 ( ) 2 1 Uf = C + C V (where ) 1 2 1 1 2 2 C C C V C V V + = Hence energy loss 2 1 2 1 2 1 2 ( ) 2( ) V V C C C C U Ui Uf − +  = − = 80. (b) The two capacitors are in parallel so ( ) 2 1 2 0 k k t A C +  =  81. (c) 2 1 2 1 2 1 1 = + + C  C F 3 2 = 82. (c) Charges developed are same so C1V1 = C2V2  2 2 1 = V V V1 + V2 = 120  V 80 volts 1 = 83. (a) Given circuit can be drawn as Equivalent capacitance = 4  8 = 32F 84. (d) The given circuit can be redrawn ass follows Equivalent capacitance of the circuit CAB = 4F Charge given by the battery Q = CeqV = 4  60 = 240 C Charge in 5F capacitor Q 240 50 C (10 5 9) 5 '  = + + = 85. (b) The given circuit can be redrawn as follows 86. (b) The given arrangement is equivalent to the parallel combination of three identical capacitors. Hence equivalent capacitance d A C 0 3 3  = = 87. (d) Total capacitance 12 1 8 1 20 1 1 = + + C  C F 31 120 = Total charge Q CV 300 1161 C 31 120 = =  = Charge, through 4 F condenser 580 C 2 1161 = = and potential difference across it 145 V 4 580 = = 88. (c) 2 2 1 U = CV Now if V is constant, then U is greatest when 'Ceq' is maximum. This is when all the three are in parallel. 89. (d) 90. (b) Equivalent capacitance of the circuit Ceq = 6F Charge supplied from source Q = 6  20 =120 C Equivalent capacitance Ceq F 2 3 1 2 1 = + =  1F 1F A B 2 1 F 1F A 8 F B 8 F 8 F 8 F 9F 60 V 5F 10F 12F 8F A B Q Q' A B A B C C C  CAB = 3C  CAB 3F 4 12 4 12 = +  = A B 2F 2F 4F 12F