Title 17. Electrostatics Easy Ans.pdf ✅

1. (d) Coulomb's law is used to calculate the force between charges. 2. (d) 2 1 r F  ; so when ris halved the force becomes four times. 3. (b) The same force will act on both bodies although their directions will be different. 4. (c) The force will still remain 2 0 1 2 4 r q q  5. (d) Gravitational force between electrons 2 2 ( ) r G m F e G = Electrostatics force between electrons 2 2 . r e F k e = 9 19 2 11 31 2 2 2 9 10 (1.6 10 ) 6.67 10 (9.1 10 ) . ( ) − − −       = = k e G m F F e e G 43 2.39 10 − =  6. (b) 2 0 1 2 2 0 1 2 4 , 4 K r q q F r q q Fa b   = =  Fa : Fb = K : 1 7. (b) Due to mutual repulsion of charges distributed on the surface of bubble. 8. (c) We put a unit positive charge at O. Resultant force due to the charge placed at A and C is zero and resultant charge due to B and D is towards D along the diagonal BD. 9. (c) Surface charge density A q  = 10. (a) Excess of electron gives the negative charge on body. 11. (c) All other charges are its integral multiple. 12. (c) Gravitational force and nuclear force both are attractive in nature. 13. (d) Q1 + Q2 = Q ..... (i) and 2 1 2 r Q Q F = k .....(ii) From (i) and (ii) 2 1 1 ( ) r kQ Q Q F − = For F to be maximum 0 1 = dQ dF  2 1 2 Q Q = Q = 14. (a) The force between 4q and q; 2 0 1 4 4 1 l q q F  =   The force between Q and q; 2 0 2 4 ( / 2) 1 l Q q F  =   We want F1 + F2 = 0 or 2 2 2 4 4 l Qq l q = −  Q = −q 15. (a) The charge given to a sphere will be distributed uniformly over the surface. 16. (a) The position of the balls in the satellite will become as shown below Thus angle  = 180° and Force 2 2 0 4 (2 ) 1 L Q =   17. (a) N r k Q F 3 2 9 2 2 2 9 10 (1000 ) 1 = = 9 10 1  =  18. (d) Resultant charges after adding the – 2C be (− 2 + 2) = 0 and (− 2 + 6) = + 4C  F 0 0 4 2 2 1 2 =  = =  r k r k Q Q 6. (a) 12 0 81 8.854 10 −  = K =   MKS units 10 7.17 10 − =  20. (c) Because in case of metallic sphere either solid or hollow, the charge will reside on the surface of the sphere. Since both spheres have same surface area, so they can hold equal maximum charge. 21. (c) For providing path to charge induced on the surface of the carriers which take inflammable material. 22. (b) 2 2 0 12 4 1 a q F =   and 2 2 0 13 4 ( 2) 1 a q F =    2 13 12 = F F 23. (c) Net force on B 2 2 Fnet = FA + FC ( ) F dyne A 20 3 15 12 2 =  = , ( ) F dyne C 15 4 12 20 2 =  =  F F F dyne net A C (20) (15) 25 2 2 2 2 = + = + = 24. (b) With temperature rise, dielectric constant of liquid decreases. 25. (d) In the presence of medium force becomes times K 1 . 26. (a) Separation between the spheres is not too large as compared to their radius so due to induction effect redistribution of charge takes place. Hence effective charge separation decreases so force increases. +Q L L +Q 180o A FC FA 3 cm 4 cm +15 esu – 20 esu +12 esu B C 2 2 net A C F = F + F
27. (a) 14 19 10 1.6 10 − Q = ne =    Q 1.6 10 C 16 C 5 =  = − Electrons are removed, so chare will be positive. 28. (a) When put 1 cm apart in air, the force between Na and Cl ions = F. When put in water, the force between Na and Cl ions K F = 29. (d) Positive charge shows the deficiency of electrons. Number of electrons 9 1.6 10 14 .4 10 19 19 =   = − − 30. (c) 31. (c) Initially, force between A and C 2 2 r Q F = k When a similar sphere B having charge +Q is kept at the mid point of line joining A and C, then Net force on B is Fnet = FA + FC ( ) ( ) F r k Q r k Q r Q k 8 8 2 2 2 2 2 2 2 2 = + = = . (Direction is shown in figure) 32. (a) Let separation between two parts be r  2 ( ) . r Q q F k q − = For F to be maximum = 0 dq dF  1 2 = q Q 33. (b) 18 19 6.25 10 1.6 10 1 =   = = − e Q n 34. (a) K F r q q F = = 2 1 2 4 0 1  If F is the force in air, then F is less than F since K  1 . 35. (b) Charge on glass rod is positive, so charge on gold leaves will also be positive. Due to X-rays, more electrons from leaves will be emitted, so leaves becomes more positive and diverge further. 36. (d) Negative charge means excess of electron which increases the mass of sphere B. 37. (b) 2 1 2 2 1 2 1           = r r F F r F  F N F 11.25 0.06 5 0.04 2 2 2  = =      = 38. (a) K F 1  i.e. K F F air medium = 39. (a) In second case, charges will be − 2C and +3C Since F  Q1Q2 i.e. 1 2 1 2 Q' Q' Q Q F F =   4 2 3 40 3 8 = − −   = F  F = 10 N (Attractive) 40. (b) By using 19 19 10 1.6 10 − Q = ne  Q =   = +1.6C . 41. (b) FA = force on C due to charge placed at A 1.8 N (10 10 ) 10 2 10 9 10 2 2 6 6 9 =    =   − − − FB = force on C due to charge placed at B 1.8N (0.1) 10 2 10 9 10 2 6 6 9 =   =   − − Net force on C F F F F F N o net ( A ) ( B ) 2 A B cos120 1.8 2 2 = + + = 42. (c) By using Q ne Q e C 19 2 3.2 10 − =  = + = +  43. (b) F  Q1Q2  1 8 5 5 10 20 ' 2 ' 1 1 2 2 1 = − −  −  − = = Q Q Q Q F F 44. (c) By using 2 2 9 9 10 . r Q F =   F 0.144 N (0.5) (2 10 ) 9 10 . 2 6 2 9 =  =  − 45. (c) Effective air separation between them becomes infinite so force becomes zero. 46. (a) 2 2 9 9 10 r Q F =   N 8 1 0 2 1 9 2 9 2.3 10 (10 ) (1.6 10 ) 9 19 − − − =   =   47. (c) Number of atoms in given mass 23 6.02 10 63.5 10 =   = 9.48  1022 r A C +Q – Q B r/2 r/2 +Q FA FC +1C – 1C +2C 10 cm FB A B C FA 120o + – A B 10 cm e –
Transfer of electron between balls 6 22 10 9.48 10 = = 9.48  1016 Hence magnitude of charge gained by each ball. Q = 9.48  1016  1.6  10–19 = 0.015 C Force of attraction between the balls 2 10 . (0.1) (0.015 ) 9 10 8 2 2 9 F =   =  N 48. (a) Surface charge density () Surface area Charge = So 2 4 2 b Q inner   − = and 2 4 c Q Outer   = 49. (a) In the following figure since | | | | | | FA = FB = FC and they are equally inclined with each other, so their resultant will be zero. 50. (d) By using 4 2.5 10 10 5 4 =  =  = − − K F F K m a 51. (c) 2 2 | | | | . a Q F F k B = C = Hence force experienced by the charge at A in the direction normal to BC is zero. 52. (d) They will not experience any force if | | | | FG = Fe  2 2 2 0 2 2 2 (16 10 ) . 4 1 (16 10 ) − −  =  m q G   G m q = 4 0 53. (b) On rubbing glass rod with silk, excess electron transferred from glass to silk. So glass rod becomes positive and silk becomes negative. 54. (c) r r e r k r e F k 3 2 2 2 = − ˆ = − .         = r r  rˆ 55. (c) By Q = Ne or 1 4 1 9 6 5 10 1.6 10 80 10 =    =  = − − N e Q N 56. (c) F = F' or K r r r K Q Q r Q Q =  ' = 4 4 ' 2 0 1 2 2 0 1 2   57. (b) Dielectric constant 0   K = Permittivity of metals () is assumed to be very high. 58. (c) Potential energy depends upon the charge at peaks of irregularities. Since every event in the universe leads to the minimisation of energy. 59. (c) Let us consider 1 ball has any type of charge. 1 and 2 must have different charges, 2 and 4 must have different charges i.e. 1 and 4 must have same charges but electrostatics attraction is also present in (1, 4) which is impossible. 60. (c) After following the guidelines mentioned above Fnet = FAC + FD = FA + FC + FD 2 2 Since 2 2 a kq FA = FC = and 2 2 (a 2) kq FD =       = + = + 2 1 2 2 2 2 2 2 2 2 2 a k q a k q a k q Fnet         + = 2 1 2 2 4 2 0 2 a q  61. (c) Since both are metals so equal amount of charge will induce on them. – Q + 2Q = Q – 2Q +2Q b c a q A C B q q FA Q FB FC 60o A +Q FB FC – Q B C a FC sin 60o FB sin 60o 60o 60 FC cos 60 o o FB cos 60o 60o 60o A D C B FC FAC FA FD +Q +Q +Q
62. (d) Initially 2 2 . r Q F = k (fig. A). Finally when a third spherical conductor comes in contact alternately with B and C then removed, so charges on B and C are Q / 2 and 3Q / 4 respectively (fig. B) Now force F r Q Q F k 8 4 3 3 2 ' . 2 =             = 63. (b) When a positively charged body connected to earth, electrons flows from earth to body and body becomes neutral. 64. (a) N r r F 2 12 0 2 6 6 0 35 10 4 ( 7 10 )( 5 10 ) 1 4 1  = − +  −  = − −   N r r F 2 12 0 2 6 6 0 35 10 4 ( 5 10 )( 7 10 ) 1 4 1 '  = − +  −  = − −   65. (a) Gravitational force 2 r Gm m F e p G = 1 1 2 1 1 3 1 2 7 (5 10 ) 6.7 10 9.1 10 1.6 10 − − − −       FG = = 3.9  10–47 N Electrostatic force 2 2 4 0 1 r e Fe  = 1 1 2 9 1 9 1 9 (5 10 ) 9 10 1.6 10 1.6 10 − − −       Fe = = 9.22  10–8 N So, 39 47 8 2.36 10 3.9 10 9.22 10 =    = − − G e F F 66. (d) F  Q1Q2  ' ' 1 2 1 2 2 1 Q Q Q Q F F = 1 4 3 2 3 8 (3 10 6 10 )(8 10 6 10 ) 3 10 8 10 6 6 6 6 6 6 = − −   =  −   −     = − − − − − −  N F F 3 3 1 2 1.5 10 4 6 10 4 − − = −   = − = − (Attractive) 67. (a) Initially 2 2 r Q F = k ....... (i) Finally Force on C due to A, 2 2 2 2 ( / 2) ( / 2) r kQ r k Q FA = = Force on C due to B, 2 2 2 2 ( / 2) ( / 2) r KQ r KQ Q FB = =  Net force on C, F r k Q Fnet = FB − FA = = 2 2 68. (d) 2 2 . r Q F = k . If Q is halved, r is doubled then times 16 1 F → 69. (b) The schematic diagram of distribution of charges on x- axis is shown in figure below : Total force acting on 1 C charge is given by       +   = − − 2 6 2 6 0 (2) 1 1 10 (1) 1 1 10 4 1  F     +    +   + − − .... (8) 1 1 10 (4) 1 1 10 2 6 2 6             −  =        = + + + +  − − 4 1 1 1 ... 9 10 10 64 1 16 1 4 1 1 1 4 10 9 6 0 6  3 4 9 10 3 4 9 10 10 9 6 3 =    =   − = 12000 N 70. (a) 19 19 10 1.6 10 1.6 =  = = − e q n 71. (a) In case of spherical metal conductor the charge quickly spreads uniformly over the entire surface because of which charges stay for longer time on the spherical surface. While in case of non-spherical surface, the charge concentration is different at different points due to which the charges do not stay on the surface for longer time. 72. (b) Nuclear force binds the protons and neutrons in the nucleus of an atom. Q B C r Q (A) Q/2 B C r 3Q/4 (B) + + + + + + + + + e – A B r Q Q A C r/2 Q/2 Q B r/2 FB FA Q/2 x =1 1C 1C 1C 1C O x =2 x =4 x =8 1C