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ELECTROSTATICS

SECTION 3 SOLUTIONS

1. Charging by conduction is addition of electrons when nega-tively charged) or removing electrons when positively charged). Here electrons are added. The mass slightly in-creases. Ans.a

2. A metal plate has infinite dielectric constant. The force be-tween charges separated partly or fully by it will be (1/4πε0εr) (q1q2/r2) with εr = α. Thus the force between them will be zero. Ans.a

3. The force between two point charges q1 and q2 kept at a dis-tance F12 and F21 will have same magnitude (since the force depends on the product of the charges) and opposite direction. i.e. F21 = -F12. Hence the ratio will 1: -1. Ans.d

q

qq

EE

E

4. The potential on the surface of a charged sphere is Aq/r. When same charges are given sphere of smaller radius will have greater potential. When connected by wire charge flows from higher to lower potential making the potential same (choice b). This re-duces charge of smaller sphere and increases charge of bigger sphere (choice d) Ans.b & d

5. Refer to the adjacent figure. The electric field at the centre of the circle 0 has a magnitude Aq/r2 due to each charge. Their directions are shown. Three such vectors form the sides of an equilateral triangle and by rules of vector addition the sum will be zero. Ans.b

r

R

6. Refer to the adjacent figure. If ρ is the charge per unit volume, ρ = q/4πR3. Charge on the smaller sphere will be q(4/3)πr3/(4/3)πR3 = qr3/R3 where r is the radius of the smaller sphere. The electric field at the point is A (charge inside the sphere)/r2. This gives field = Aqr3/R3r2 i.e. α to r. Ans.d

7. According to Gauss’s theorem total flux crossing normally through the whole cube = q/ε0. This is flux coming out of all the six faces of the cube. Hence flux through one face = q/6ε0. Ans.d

8. When a conductor is charged, there is an outward mechanical force given by (σ2/2ε0)N/m2, where σ is charge radius of the bubble. It goes on expanding until surface tension of soap solution balances this force and brings about an equilibrium. Ans.b

9. The capacitance of a spherical conductor is 4πε0r. When connected by a wire the final potential = V = q1`+q2/c1+c2 = (1 x 10-2 + 5 x 10-2)/4πε0x3x10-2 = 2/4πε0 volt. Charge of smaller sphere = potential x capacity = (2/4πε0) x 4πε0 (1 x 10-2 )= 2 x 10-2 C. Ans.a

10. Work W = Fscosθ, where F is force, s is distance θ is angle between force and displacement. Here F = qE, where q is charge E is electric field. Thus W = qE cosθ. That is 4 = 0.2 x E x 2 x cos600. This gives E = 20 N/C. Ans.d

11. If v is the velocity of a particle of mass m, falling under a potential difference of V volt, electrostatic potential energy eV = kinetic energy (1/2) mv2 v = ⇒ 2 m/s. Ans.a

12. Unit of ε0 is farad/metre, capacitance per unit length. This can be easily checked from the equation of capacitance of a parallel plate condenser, C = ε0A/d, where a is area of the plates and d distance be-tween the plates. ε0 = Cd/A. Ans.b

13. At ground state of hydrogen atom, the centre of mass of the revolving electron and that of the nucleus will be same point that is centre of atom. Hence the separation will be zero. The dipole moment is the product of charge and separation. Dipole moment P = qa = qx0 =0. Ans.c

14. The resulting potential is total charge/total capacitance (q1+q2)/(C1+C2) =(C1V1+C2V2)/(C1+C2) = 133 V. Ans.c

15. Dielectric strength is the minimum value of the electric field to break down the specimen. The corre-sponding potential difference is field x distance = 106 Vm-1 x 10-3 m = 1000 V. Ans.c

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Physics for IIT-JEE Screening Test 216. Charge q is already in equilibrium. Consider the equilibrium of one of the Qs.This will be in equilib-

rium if force between Q and Q is equal and opposite to force between Q and q. If we take distance be-tween Q and Q as 2a, distance between Q and q is a. Then, we have AQq/a2 = -A(QxQ)/(2a)2. This will solve to q = -Q/4. Ans.b

17. When the charging battery is disconnected, the charge on the capacitor remains constant. Hence an-swer (a) is wrong. When plates are separated more the capacitance of the condenser ε0εrA/d decreases as d increases. So answer (c) is wrong. Q = V/C. Q is constant. C decreases. Hence V increases (b) is correct. The electrostatic energy of a condenser is Q2/2C. Q is constant C decreases and therefore energy increases. (d) is correct. Ans.b & d

18. Let Q be the charge and q be its fraction. Then the charges are q and Q-q. The force between charges, F = [A(Q-q)q]/r2. The force is maximum when dF/dq = 0. This gives q = Q/2. Hence force is maxi-mum between two charges when kept at a given distance if they are divided in the ratio 1:1. Ans.c

19. Since battery connection exists, potential difference V is constant. Distance between the plates d is constant . Electric field V/d, therefore, is constant. Ans.c

20. Energy gained by the electron = q x V, where q is charge and V is potential difference. Since electric field E is uniform, V = Exd, d separation between the plates. Therefore energy = qEd = 1.6 x 10-19 x 2.5 x 2 = 8 x 10-19 J. Ans.d

21. The work done by the battery to charge the condenser is qV J. However, the energy stored by the condenser is only (1/2)qV. The factor 1/2 appears here because during charging potential increases uniformly from 0 to V. Hence we take the average value. Ans.b

22. The field due to a dipole on the axial line at a distance is Ax 2p/r3. Since field is potential gradient, potential should be proportional to 1/r2. Hence when r is doubled, V becomes V/4. Ans.b

23. At a point electric potential or field can exist even when the other is zero. Examples are given in indi-rect theory notes. Ans.c, d.

24. The number of lines of force coming out from a unit charge in CGS system is 4π and iun SI = 1/ε0. Value of ε0 = 8.85 x 1012. Hence 1/ε0 = 1/(9 x 10-12) = 1012/9 = 1.11 x 1011. This is also given in indi-rect theory notes. Ans.a

25. Two plates get equal potential V and charge Q = CV. They are joined in series. No charge flows as they carry the same potential, and in a series connection the charge is same. Potential difference = 2V. Ans.d

26. When the distance is increased capacitance = ε0ε0A/d decreases. Since charge is the same, the potential increases by the equation, Q = CV. Ans.c

27. If q is the maximum charge it can hold, q/4πε0r2 is the electric field on the surface. q = 4πε0r2 E. Here r = 5/2 m. 4πε0 = (9 x 109)-1. We have from this q = 2 x 10-3 C. Ans.b

28. The field inside a charge hollow sphere is zero. Therefore the potential at all points will be same in-side. Ans.a

29. Force of attraction (F1) = A(10) x (-20)/r2 = -200A/r2 where A is the constant appearing in Coulomb’s law. When the balls touch each other, the radii being the same, the capacitance will also be the same. So the charge will distribute equally. The charge q1 = q2 = (10-20)/2 = -5 units. (F2) = A(-5) x (-5)/r2 = 25 A/r2 , F1:F2 = -8:1. Ans.a

30. Taking 1/4πε0= A, the potential at the surface of larger sphere AQ1/R1 should be = potential inside at a distance r from the centre because then only the electric field inside larger sphere will be zero. The potential due to smaller sphere will be AQ2/r, assuming charge concentrated at centre. So the total po-tential = (AQ1/R1) + (AQ2/r). Ans.c

31. When charging battery is connected V remains constant. When distance is increased, electric field V/d dcreases.. Ans.b.

32. Using the formula C = ε0Ak1k2 / (k1t2 + k2t1), for a parallel plate condenser filled with 2 dielectrics, we have here t1 = t2 = t/2, where is total distance between the plates. This gives C = 2ε0Ak1k2/(k1+k2)t. Ans.c

33. The electric field between parallel plate condenser is uniform and is given by E = σ/ε0, from Gauss’s law. Force = charge x field. Hence the force on all the protons will be same in magnitude and direc-tion. Ans.a,c

34. Since the field increasing along x-axis force on –q is greater than force on +q. Hence the dipole will move along direction of greater force i.e. negative x-axis. The force on charge will produce a torque which rotates the dipole anti-clockwise. Ans.d

35. Taking 1/4πε0 = A, we have Aq1/a = Aq2/b. That is q1/q2 = a/b ....(1) If σ1 and σ2 are ratio of surface density of the two spheres, σ1 / σ2 = (q1 /4πa2)/(q24πb2) = (q1/q2)(b2/a2). Substituting from (1) for q1/q2, we get σ1/σ2 = b/a. Ans.a

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Electrostatics 3

36. .k200E = The area vector S is perpendicular to the plane of the coil that is along the z axis. There-

fore S = 10 x 20 x 10-4 S.EfluxThe.k =φ = 2000 x 200 x 10-4 ( = 40 units. Ans.b )k.k37. The energy of a charged conductor is given by (1/2) qV or (1/2) CV2. For a spherical conductor V =

q/4πεoR. Therefore energy will be q2/8πεoR. Ans.d 38. The intensity of electric field at a point is the number of lines of force crossing unit area around the

point normally. Here it is 10. Ans.d 39. Sd.E=φ E dS cosθ, where θ is the angle between field vector and area vector. Here the lines make

an angle 300 with the surface, which means they make an angle 600 with the area vector. (Recall area vector is normal to the surface). Thus φ = 10 x 1 x cos600 = 5. Ans.c

40. Since the coil is Y-Z plane, its area vector will be 200 cmi 2. φ = E. dS = (200 k ). (200 x 10-4 =0,

because = 0. Ans.a

)i

i.k41. Inside a satellite there is no other force except electrostatic force (apparent weight of the balls is zero).

Therefore due to the electrostatic force of repulsion the balls settle at maximum distance. Both the strings will be horizontal. Angle between them will be 1800 . Ans.c

42. The tension of each string will be equal to the electrostatic force of repulsion experienced by a ball.

The distance between the balls is 2L. Therefore the force =⎥⎥⎦

⎤

⎢⎢⎣

⎡

πε 20 )L2(

QxQ4

1 Ans.b

43. Force on electron due to electric field = eE. If this is equal to weight of electron, we have mg = eE. E = mg/e. Ans.b

44. Taking A = 1/4πεo, potential on this surface = Aq/r. Potential at a point 3r from the centre Aq/3r. It is given (Aq/r)-(Aq/3r) = V. That is V = (2/3)Aq/r. field at 3r = -d/dr = (Aq/3r) = Aq/3r2 = V/2r. Ans.d

45. Refer theory notes and previous similar question, in which it has been shown that the electric field in-side a charged conducting sphere is proportional to distance from the centre. Therefore field decreases as we penetrate from the surface towards the centre. Ans.a

46. 1 g of hydrogen 6 x 1023 atoms nearly. If all electrons are removed it will have 6 x 1023 protons left. The total charge of all these protons will be 1.6 x 10-19 x 6 x 1023 = 9.6 x 104 ≈ 105 C. Ans.d

47. Electric field E = -dV/dx. That is E = -(10x+10). At the point x = 1m, E will be equal to –(10x1+10) = -20 V/m. Ans. d

48. At a point on the perpendicular bisector of a dipole, the charges are at equal distance. Hence the electric potential will be A(q/r) + A(-q/r) = 0. That is (b) is correct. The electric field at this point (on the equitorial line) will be Ap/r3, and this always be parallel to the axis of the dipole. (d) is correct. Ans.b & d. (Note A = 1/4πεo)

49. Since the battery is disconnected, charge Q on the condenser remains the same. The capacitance of the condenser when the slab is inserted increases to K times. Work done = difference in the electrostatic energy = (Q2/2C)-(Q2/2CK) = Q2(K-1)/2KC. Since Q = CV and C = εoA/d, this reduces to εoAV2(K-1)/2dK. Ans.c

50. If both E and B are zero, the proton will certainly continue with constant velocity. (a) is correct. Even when B is not equal to zero, if it is parallel to proton motion, it will go undflected. Hence (b) is cor-rect. If E is not equal to zero proton will be accelerated (c) is wrong. If both E and B exist, and the force due to the fields are equal and opposite i.e. eE = evB, the proton would move with constant ve-locity v = E/B. (d) is correct. Ans. a,b & d

51. When the positive charge is released, the x-components 2Fx of coulomb force F will provide restoring force for oscillatory motion, while y components cancel. However, coulomb force obeys inverse squa-relaw, while we need a force proportional to displacement for SHM. Hence the charge will oscillate but not simple harmonically. Ans.d

52. If n drops of same radius and charge coalesce, the potential of resulting drop V1 can be found by the formula V1 = n2/3 V ( Indirect theory notes), where V is the potential of smaller drop. The required ra-tio, therefore, V1 /V is (1000)2/3 = 100. Ans.c

53. Taking A = 1/4πεo, the potential at the origin will be V(0) = (Ae/1)+(Ae/2)+(Ae/4)+(Ae/8) +.....to in-finity. This is equal to Ae[1+(1/2)+(1/4)+(1/8)....]= 2e/4πεo, where we observe the sum of geometric series in the bracket is 2. Ans.b

54. In this case V(0) = (Ae/1) – (Ae/2) + (Ae/4)- (Ae/8)+.....= Ae[(1)-(1/2)+(1/4)-(1/8)....]= Ae(2/3) = e/6πεo, where we observe the sum of geometric series in the bracket is 2/3. Ans.d

55. The electric field at the origin E(0) = (Ae/12)+(Ae/22) + (Ae/42) + (Ae/82) ...= Ae[1+(1/22) + (1/32) + (Ae/42)-(Ae/82)+......= Ae[1+(1/22) + (1/42) +.......] = 3/4πεo, where we observe the sum of geometric series in the bracket is 4/3. Ans.c

56. The electric field E(0) = (Ae/12) – (Ae/22) + (Ae/42)-(Ae/82) +....= Ae[1-(1/22)+ (1/42).....] = Ae(4/5), where we observe the sum of the geometric series in the bracket is 4/5. Hence E(0) = e/5πεo. Ans.d

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Physics for IIT-JEE Screening Test 457. Since charge is same we use equation for energy (1/2)q2/C. This means energy U α 1/C. The ratio of

energy will be U1/U2 = C2/C1 = 0.6/0.3 = 2. Ans.d 58. The field inside a charged hollow conductor is zero. This will be satisfied only if the potential differ-

ence remains the same value V. Ans.a 59. In order to measure electric field, we place a test charge q0. The test charge will have to be as small as

possible because otherwise it will produce its own field at the point and thus will disturb the electric field to be measured. Hence the measured field will always be less than the actual field. Ans.c

60. The electrostatic energy lost is the heat energy produced. Hence (1/2) CV2 is the heat energy pro-duced. Hence C = 2 x 10-6 F. V = 200 volt. We get energy = 0.04 J. Ans.c

61. When glass plate is fully inserted, capacitance increases to εr times, where εr is dielectric constant of glass. Since charge is constant potential will decrease to 1/εr times. Here 1/εr is 1/8. Therefore εr = 8. Ans.b

62. Electric field between parallel plate condenser is uniform and is σ/ε0, where σ is charge per unit area. When one plate is removed, the field becomes that due to one charged plate. The field will be either σ/2ε0 or - σ/2ε0. Force on a charged particle = field x charge. Hence the force will reduce to half. Ans.b

63. Since they are connected to the same source, V is constant. Use the equation for energy (1/2)CV2 . In the first case capacitance is C/2 and in the second case 2C. So the ratio of energy will be 1: 4. Ans.a

64. When the spheres are immersed in liquid due to buoyant force their apparent weight will be less. Hence net downward force will decrease (c) is correct. Electrostatic force is balanced by the horizon-tal component of tension. Hence (b) is correct. Ans.b & c

65. Vandegraff generator produces very high electric potential of the order of one million volt so that charged particles can be accelerated using its potential. Ans.d

66. Potential energy of a system of three given point charges = (-Aq2/r) + (-Aq2/r) + (Aq2/r) = -Aq2/r. Ans.d

67. A uniform electric field exists between the plates. A metal sphere suspended as shown when slightly displaced will move towards one plate. Due to restoring force produced by the string it moves in the opposite direction. Thus it will oscillate with a constant period. Ans.c

68. The electrostatic energy gained by A and B will be respectively qV, 4qV. This is equal to respective mechanical kinetic energies (1/2) mvA

2, (1/2)mvB2 . From this we get vA

2 / vB2 = 1/4. vA / vB is 1:2.

Ans.b 69. At the ground state, potential energy of hydrogen atom is –27.2 eV. Its kinetic energy (half of poten-

tial energy) is 13.6 eV, which makes the total energy as –13.6 eV. Since electrostatic potential energy of electron is –27.2 eV, electrostatic potential at the site of electron due to proton is +27.2 V (+ as its is due to proton). Ans.a

70. The effective capacitance between B and C is 3µF. (Balanced Wheatstone’s bridge of capacitors). This is in series with 3 µF between A and B and 3 µF between C and D. These three 3 µF capacitors in series add to 1 µF by series addition rule. Ans.d

71. Here weight mg = qE. Therefore q = mg/E = 8 x 10-3 x 10/1000 x 103 = 8 x 10-9 C. The number of quantum charges = q/charge of electron = 8 x 109 / 1.6 x 10-19 = 5 x 1011. Ans.a

72. If n is the number of electrons carried by each and e is the electron charge, we have A(ne)2/I2 = 2.56x10-9 N by Coloumb’s law, where A = 1/4πεo = 9 x 109 . This simplifies to n2 = 1020/9. n = 1010/3 = 3.33 x 109. Ans.a. (Note: The numbers given here are easy for simplification)

73. Volt x coulomb = potential x charge = energy, which has the same dimension as torque. Ans.d 74. When electric field is perpendicular to the direction of motion of electron with uniform speed, electron

will have velocity and acceleration perpendicular at the beginning. Hence its path will be a parabola in the plane of velocity (X) and acceleration (Z). Ans.c

75. When applied magnetic field is along Z direction and velocity along X-direction by Lorentz force equation q v x B the force will be along Y direction. So electron will move in a circle containing ve-locity vector and force vector. That is X-Y plane. Ans.d

76. Here we have two plates forming a parallel plate condenser. The electrostatic energy between the plates = Aq(-q)/r = -Aq2/r. Ans.d

77. The field due to charged plane sheet = σ/2εo, which is uniform and does not depend on distance. So it will be same at infinity. Ans.a

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Electrostatics 5 78. If θ is angle made by the string with vertical, the forces acting on the bob are mg vertically

downward and qE horizontal. The string will settle at an angle θ, such that tanθ = horizontal force/vertical force = qE/mg. (Recall tangent law in dynamics). Ans.b

79. The electric field due to a cylinder = λ/2πε0r, where λ is charge per unit length. The field is propor-tional to 1/r. The graph between two quantities inversely proportional to each other will be a rectangu-lar hyperbola. Ans.c

80. Acceleration = qE/m. Velocity after t seconds v = vo+at = 0 + (qE/m)t. Kinetic energy = (1/2) mv2 = (1/2)m(qEt/m)2 = (1/2)q2E2t2/m. Ans.c

81. According to Columb’s law if r is distance in air and r1 is a distance in the medium for the forces to be equal r2 should be equal to εrr1

2. Equating we get r1 = r/√εr = r/2. Ans.b 82. Potential V = Aq/r, where A = 1/4πεo. Here q is the same. Hence V α 1/r. V1/V2 = r2/r1 = 3/5. Ans.a 83. Here V is same. Using V = Aq/r, q α r. q1 /q2 = r1 / r2 = 5/3. Ans.b 84. Electric field at the surface = Aq/r2, where A = 1/4πεo. Here q is same. Hence E α 1/r2. E1/E2 = r2

2 /r1

2 = 9/25. Ans.c 85. Since electric field in the surfaces are same , Aq/r1

2 = Aq2/r22. This gives q1/q2 =25/9. Ans.d

86. The force between q1 and q2 depends only on q1 and q2 and their separation according to Coloumb’s law. Any intervening charge q3 will not change the force between these two. Ans.b

87. With twelve condenser plates we can form eleven condensers. If the plates are arranged parallel at equal distance and alternate plates are connected to positive of a battery, other alternate plates to nega-tive, these condensers will be parallel. Its capacity will be 11 µF. Ans.d

88. Taking A = 1/4πεo, potential due to charge at the adjacent corner = Aq/a. Potential due to charge in the opposite corner = Aq/√2a. Potential difference = Aq/a [1-(1/√2)]= Aq/a[1-0.71] = 0.29 x Aq/a, where we have taken 1/√2 = 0.71. Ans.d

89. Potential on the surface of A is = potential on the surface of B, because the ratio 10/3 = 20/6 (potential = 1/4 πεo(q/r). Since there is no potential difference charge will not flow. Ans.c

90. Use the information supplied already in theory notes.. This law fails inside the nucleus. The inter-nuclear distance is of the order of 10-15m. The nuclear forces follow a different set of laws. Ans.d

91. An electric charge will move along the line of force only if it is free, i.e. it is not under the action of any other force. Hence it cannot move along the line of force if it is accelerated. If its velocity is uni-form, still it need not follow the line of force if its velocity is inclined to the field. Ans.a

92. The dielectric constant of pure water is 81. Among the given media water has maximum dielectric constant. Since the coulomb force is inversely proportional to dielectric constant εr , it is minimum for water. Ans a

93. Use the information supplied already in theory notes.. The force between q1 and q2 will not change due to the presence or absence of the third charge q3.. Ans.c

94. At the centre the electric potential due to the charges will be 14

2πε o

qr

qr

qr

( − − ), where r is the dis-

tance from the corner to the centre of the triangle. The electric field at the centre will not be zero, be-cause the fields are in different directions due to charges and hence will not cancel. Ans.b

95. In an electric field the line of force is the path followed by an isolated (or free) charge. Initially the charge is in the line of force because x = 1, y= 0 is a point in the circle. Hence when it is free, it will move along the line of force. Ans.d.

96. Use the information supplied in theory notes.. The total number of lines of force coming from the charge is 1.11x 1011. This is distributed to six faces. The number of lines of force coming from one face = (1/6) 1.11x 1011. Ans.d.

97. The force on 1C is 5 N means electric field at the origin is (force / charge) = 5/1 = 5 volt/m. Since the field increases at a rate 1 V/m for every metre, the field at x = 1 should be 5+1 = 6 V/m. The force on -1 C = charge x field = -1x6 = -6 N. Ans.b

98. Use information supplied in theory notes. The electric field due to a charged plane sheet ( σ/2εo ) is same at all points. The field is uniform throughout. Ans.d..

99. Use the information supplied already in theory notes. When the distance increases to twice the field decreases to 1/23 of E. i.e. E/8. Ans.c

100. The field inside a charged hollow conductor is zero. In order to make the field zero, the charges will have to flow from inner hollow sphere Q to the outer one. Ans.c

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Physics for IIT-JEE Screening Test 6he

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101. T potential due to a point charge q is given by V = q/ 4πεor. The field due to a point charge at the same point will have a magnitude E = q/4πεor2. From these two equations, we find E = V/r = 6/2 = 3 N/C. Ans.a

102. W n a dipole is disturbed slightly, there will be a restoring torque trying it bring back to equilibrium. Due to the action of the torque, it returns. But it would have acquired kinetic energy of rotation, when it returns to equilibrium. Therefore, it continues in the opposite direction. Thus it oscillates. When all its energy is spent against dissipative forces, it returns to equilibrium. Ans.c

103. The two torques acting on the dipole are pFsinθ and p2F sin(90-θ), where θ is angle made by the di-pole with X axis at equilibrium. When the two torques are equal, the dipole will remain in equilibrium. Equating, we get tanθ = 2. Ans.c

104. T particle here is subjected to an electric field perpendicular to its velocity, because velocity is along Z-axis and the field along X axis. Hence the path will be a parabola in the plane of velocity and the field vectors. Ans.b

105. According to Lorentz’s force equation the force on the particle will be F qv= the vector product of velocity and field will be perpendicular to them, the force vector will be along the Y axis. This will provide centripetal force for circular motion. Therefore, the particle will have a velocity along Z-axis and centripetal acceleration along the Y-axis. The particle will move in a circle in Y-Z plane. Ans.b

xBz x . Since

106. The work done is the total potential energy of the system. If each charge is q, then the total energy will be 3Aq2/r. Here A = 1/4πεo = 9x109. q = (1/3 µC) = (10-6 / 3) C. This gives work as equal to 3 m J. Ans.b

107. The electric field on the surface of a sphere charged with q is q/4πεor2, where r is radius of the sphere. This can be written as σ/εo, because σ = q/4πr2. Since σ is same, the electric field will also be same. Ans.a

108. The cube is placed so that the electric field is parallel to one of the edges along the X-axis. . The num-ber of lines of force entering the face perpendicular to the edge is equal to the number of lines of force leaving. Hence the net flux will be zero. Ans.d

109. It is essential that a good insulator should have high value for dielectric constant or relative permittiv-ity. The dielectric strength is the electric field to produce break down in the substance. This also will have to be high. Ans.d

110. The potential at the points A and B due to the point charge kept at the centre are same. This means potential difference is zero. Hence work done when a charge is moved between the points will be zero. You can also use theory note short cuts given. Ans.d

111. If we imagine a sphere of radius r inside bigger sphere, the question asks to find the field at the sur-face of this smaller sphere. The electric field at the surface of this smaller sphere is produced by the charge inside it. Since volume distribution of charge is uniform, the charge inside the smaller sphere will be (4/3)πr3ρ. Let this charge be q1. The electric field will be q1/ 4πεor2. Substituting for q1, we

get the field at the surface of smaller sphere as r

o

ρε3

. Ans.b

112. If the surface density of charge is σ, there will be an outward force on the soap bubble = (σ2/2εo) N/m2 due to charging. This force will always expand the soap bubble, whether σ is positive or negative. So the soap bubble will always expand when charged. Ans.a.

113. The resulting potential is total charge/total capacitance. Since the positive plate of one is connected to the negative plate of the other, the charges cancel, i.e. total charge will be q1-q2. Thus V=

021

10022001

21

2211

21

21 =+−

=+−

=+− xx

CCVCVC

CCqq

. Ans.d

114. T electrostatic potential energy = charge x potential = qV. This will be in joule. If we write it in terms of electron charge e, then the charge carried by the α particle q = 2e. The energy will be 2eV joule = 2V electron volt = 2 x1 electron volt. Ans.b

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ny115. A point on the y-axis is at equal distance from both the charges. Hence the potential due to them will be equal in value and opposite in sign, which makes (a) correct. Any point on the y-axis is a point on the equitorial line of the dipole. Hence the field will depend on 1/y3. (Look up Theory notes sup-plied). So (b) and (d) are wrong. The electric field at all points on the y-axis will be in a direction par-allel to the axis of the dipole, in this case +x axis. This makes (c) correct. Ans.a,c

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Electrostatics 7

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nc

se

116. The magnitude of the torque is given by pE sinθ, where θ is the angle made by the dipole with the

field. When the angle is made three times the torque becomes twice only if θ= 30o, since sin 30o = 0.5 and sin 90o = 1. Ans.d

117. U the information supplied in theory notes.. When three identical condensers are joined in parallel the effective value is 3C and in series the effective value is C/3. It is given here 3C-(C/3) = 16. Solving C = 6µF. Ans.b

118. Use the information supplied in theory notes. The half of the energy spent by the battery is used to charge the condenser, while the other half goes as heat. Therefore, heat produced is 2 mJ. Ans.c

119. W n capacitors are in parallel, the charge is nCV. When they are in series, the charge is (C/n)V1, where V1 is the new potential. Since charge is constant, we have nCV = CV1/n. This gives V1 = n2V. You can also use theory notes to get this straight. Ans.d

120. T electrostatic energy of a charged body E = (1/2)QV. When this body is immersed in a dielectric, its capacity becomes KC and potential becomes V/K. [Use information given in theory notes tables]. The new energy will be E1 = (1/2)QV/K = E/K. Ans.a

121. W n X-rays fall on the metal ball, some of the electrons are removed from the metal due to ionisa-tion. This produces a net positive charge on the metal ball. Due to this the ball will deflect towards negative plate. Ans.b

122. Use the information supplied in theory notes. Charge flows until the potentials are equal i.e. Q1/4πεoR1 = Q2/4πεoR2. That is Q1/R1 = Q2/R2 or Q1R2 = Q2R1. Ans.c

123. Use the information supplied in theory notes. Here the potential is reduced from 100 to 25 V. This means the capacitance has been increased to 4 times i.e. from 5 to 20 due to the connection. So in-crease in capacitance is 20-5 = 15 µF. Ans.d

124. Si e charge is conserved, use equation for electrostatic energy E = QV/2. V decreases from 100 to 25 V.i.e. to 1/4 th. So energy also decreases to 1/4. Ans.a

125. Use the equation for energy U = Q2/2C. When plates are separated to 3 times capacitance becomes C/3. The energy becomes 3Q2/2C = 3U. Thus the energy increases from U to 3U. The work done = increase in energy = 2U. Ans.b

126. U the information supplied in theory notes. If a sphere of radius R holds 1 C of charge the electric field on its surface should not exceed 3x106 N/C. This means Q/ 4πεoR2 = 3x106V/m. That is, 9x109/R2=3x106, which gives R = 3000 , which is nearly 50 m. Ans. c

127. U the information supplied already in theory notes. The resulting capacity is equal to 1000se

su

sin

1/3x 2 = 10x2 = 20µF. Ans.a

128. Use the information supplied in indirect theory notes. The potential of resulting drop is equal to 10002/3x100 = 100x100 = 10000 V. Ans.c

129. U ally in international system, units are bigger than those in in CGS system. The units in CGS system are stat coulomb for charge, stat volt for potential and stat farad for capacitance and erg for energy. 1C = 3x109stat -coulomb. 1F = 9x1011 stat farad. 1 J = 107 erg. 1V = (1/300) stat volt. From these we find the CGS unit of potential stat volt = 300 V i.e. bigger than volt. Ans.d

130. U g the equationc C1 = ε 0

1 1 2 2 3 3

At k( /

th air C t k t k) ( / ) ( / )+ +

. WitA0ε re t. He

nc

h

he

he

1=t2=t3=t/3. k1=2.

k2=3 k3=4. Substituting these values we get C1= (36/13)C. Ans.a 131. Si e the potential is same we use equation for energy (1/2) CV2. So E1/E2 =C1/C2. C1/C2.= 1/4 from

indirect theory notes. Thus the ratio E1/E2 = 1/4. Ans.b 132. W en the battery is disconnected the charge remains constant. Hence use the equation for energy as

Q2/2KC with dielectric and Q2/2C without the dielectric . Thus energy increases in the ratio 1/K:1 i.e.1/2:1 That is,the energy increases by 100%. Ans.d.

133. W n an electric field is applied to a dielectric it becomes polarised. Polarisation produces dipoles. These dipoles produce a field at any point inside which opposes the external field. Hence the field at any point will be E0 - the field produced by the dipoles i.e. it will be less than E0. Ans.b

134. W n an uncharged plate is brought from infinity near to a positively charged plate, the potential of positive plate decreases and its capacity increases. (Principle of a condenser). So when this plate is removed capacity decreases and potential increases. So (a) and (d) are correct, (b) and (c) are wrong. Ans.a,d.

NKB.CO.IN

Physics for IIT-JEE Screening Test 8

se

nc

135. U the information supplied already in theory notes. If Q is the charge of each drop and V is the po-

tential, when drops coalesce the charge becomes 10 Q and potential becomes 102/3V. Since energy is proportional to the product of Q and V, it will become 10x102/3 = 105/3 of that of one drop. You can also use theory notes short-cuts. Ans.c

136. Si e all the capacitors are in parallel total capaci tance C is equal to

F8....81

41

21

14to....161

81

41

21

124 µ=⎥⎦

⎤⎢⎣

⎡ +++=∞++++++ . Ans.d

where we have used the information that geometric series in the bracket has a sum 2. Ans.d 137. The arrangement in the triangle part ABC is a bal-

anced Wheatstone’s bridge of capacitors. The effec-tive capacitance between the points A and B of this bridge is 1µF. This is parallel with 1µF across DE. Therefore the total capacitance 1+1 =2µF. Ans.a

138. L a charge of 6q flow from the battery. It divides into three equal values 2q each as shown in figure. Each of the 2q divides to two condensers as q each. Then these charges rejoin as 6q. Taking any closed path and adding potential difference across each ca-pacitor, we have 2q/C + q/C+2q/C = 6q/C

et

1, where C1 is the equivalent capacitance of A and B.⇒C1= (6/5) C. Ans.b

139. Let X be the capacitance between A and B. (fig below). In this infinite network we have repeating units. Here one unit is a parallel an series capacitor. Break the circuit after one such unit. We find the circuit will have the same structure as the original network. Hence the capaci-tance between E and F will also be X itself. X across EF is in series with 2 . This gives 2X/2+X. This is in parallel with 2 across CD. This will give (2X/2+X)+2. This should be equal to X, i.e. capacitance

across AB. Therefore, we have 2

22X

X++ = his equation we gets X = 3.2µF. Ans.c

. Before

X . Solving t

140 introducing the metal, let the capacitance be C. The C = εoA/d. If a dielectric of thickness t and dielectric constant K is introduced the new capacitance C1 is given by

3/dtandKHere)

K11(td

AC 0 =∞=

ε= . C = 1

−−1 100x

23C3A3 0 ==

2d2ε

= 150 pF. Ans.b

141. The potential, the capacity and energy depend on the shape and the charge of the conductor. Hence

142 tic energy lost is equal to heat en-they will not be same even if they have same volume. Ans.d

. Use the information supplied already in theory notes. The electrosta

ergy gained. Therefore the heat energy produced = 12

2 2 10 200 100 16 2x x J− =− ( ) . 2 2 200+

143. In a series connection capacitors get same charge. Using the equation Q =VC, we find V is propor-

144. by (1/2) εoεrE2. Ans.d he oppo-

146 be force of attraction towards the ring. The horizontal compo-

147. positively charged body loses

tional to 1/C. V1:V2:V3 = (1/1): (1/2): (1/3) = 6:3:2. So all we have do is to divide 110 volt in this ratio. The potentials will be 60V, 30V and 20V respectively. The equation for energy density i.e. energy per unit volume is given

145. W n two charges are placed at a distance, the force on them will be the same in magnitude andsite in direction. Force on q = field x charge = Eq. Force on -3q = Eq. Field at the site of -3q = force/charge = = qE/-3q = -E/3. Ans.b

. When the charge is released there will nent of the force along the axis will provide the restoring force. This force will always be directed to-wards the centre of the ring. The particle, thus, will have oscillatory motion. But the coloumb forces follow inverse square law. For simple harmonic motion we need force proportional to displacement. Hence the motion will not be simple harmonic. Ans.a A negatively charged body gains electrons. So its mass increases. Aelectrons and its mass decreases. Hence negatively charged sphere has more mass and positively charged one has less mass. Ans.b

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Electrostatics 9

48. Use the equation for capacitance of a parallel plate condenser with a dielectric of thickness t intro-

1

duced . If C is the capacitance when medium is air we have C = εoA/d.

.tandKHere.A

C 001 =∞==

)K

(td 11−−

ε This gives C1 = C. Ans.a

149. A condenser discharges according to the equation q = qoe-t/RC, where R is the value of the resistor in

150. he same. he

151 nce of the other

152

series with C. Current i = dq/dt = -qoe-t/RC(1/RC). Since q0/C is constant equal to charging potential V, we find, the current i is not zero at time t = 0,but is constant equal to q0/CR. Therefore (a) and (c) are wrong but (b) is correct. Also dq/dt is more when C is less, from the above equation. This means condesner of smaller capacitance loses charge more quickly. Hence (d) is correct. Ans.,b, d Since the condenser is connected to the battery, whatever may happen, the potential remains tW n a dielectric is introduced, the capacity will increase to K times. Ans. d.

. The electromagnetic force is transferred through photons. A charge feels the presewhen photons are exchanged between them. Hence the time taken will be the time for photons to travel the distance between them which will be about 10-8s. This time is small but finite. Ans.c

. Use the information supplied in theory notes . The insulating property of air breaks down when the electric field ≈ 3x106V/m. The corresponding energy density is

= 1 1 9 10 3 10 402 12 6 2 3ε o E x x x x J m= =− ( ) / . Ans.c

Since the charges are of opposite sign no null point will 2 2

153. be got between the charges. So R is not a null

154. negative charges. Hence it may cancel at a

155 t

156 charge on

157 theory notes.

158y of 1/4 µF. Connect such three

159 groups and

point. Since the field will have to be equal in magnitude, the point will have to be nearer to the smaller charge. Hence a possible null point is P. Ans.a The potential at any point is produced by positive and point between them, or at a point nearer to the smaller charge. Since potential is a scalar, we need not look for the direction. Hence the two points P and R could have zero potential. Note in these ques-tions we find only location of points and not the distance. Ans.d

. The solution to this question is similar to the one given in indirectheory notes for resistors. Here one repeating unit is a series and a parallel capacitor. Break the circuit after one unit and replace it with X, as shown in adjacent figure. We get 2+X and 2 in series which is equal to 2(2+X)/2+2+X. Thus we have 2(2+X)/2+2+X = X. Solving we get X= 1.2 µF. Ans.d

. At centre of one ring, say P, the potential V due to the it is V = q/4πεoR. At the same point the elements of the other ring are at a distance √2R. Therefore the potential at P due to the other ring is q/(4πεo√2R) = V/√2. Total potential is equal to V+V/√2 = (√2+1)V/√2. Ans.d.

. Use the information supplied inEach condenser should have a capacitance 4 µF. When one of them is removed the other three are in series. They will have a capaci-tance 4/3 µF. This in parallel with 4 µF will give a net capacitance 16/3 µF. Ans.c

. Connect 4 condensers in series. It will have a capacitgroups in parallel. Total capacity will be 3x(1/4) = 3/4 µF. This is shown in the figure below.

. Connect six of them in series. The effective capacitance will be 1/6 µF. Make two such connect both groups in parallel. This will give 2x1/6= 1/3 µF