CBSE Class 12 Physics Electrostatics Objective Questions

CBSE Class 12 Physics Electrostatics Objective Questions. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

CLASS 12 PHYSICS : ELECTROSTATICS
OBJECTIVE QUESTIONS

1. Two idenical rings, each of radius R, are co-axially placed. The distance between their centres is R. A charge Q is placed on each ring. The work done in moving a small charge q from the center of one ring to that of the other is
(a) \( \frac{1}{4π ε_0} \)[\( \frac{\sqrt{2} q Q}{R} \)]
(b) \( \frac{1}{4π ε_0} \)[\( \frac{(\sqrt{2} - 1) q Q}{\sqrt{2} R} \)]
(c) \( \frac{1}{4π ε_0} \)[\( \frac{\sqrt{2} q Q}{(\sqrt{2} - 1) R} \)]
(d) zero

Answer: (d) zero

1. The potential at the centre of one ring due to both rings = \( \frac{kQ}{R} \) + \( \frac{kQ}{\sqrt{2} R} \).
2. The two rings are identical and the arrangement is symmetric, so the potential at the centre of the other ring has the same value.
3. Work done W = q (V₂ - V₁) = 0.
4. Hence, the correct option is (d).

Teacher's Note:
a) Work done in moving a charge depends only on the potential difference between the two points.
b) Symmetry removes the need for any calculation of the potential.

2. Two concentric hollow spheres of radii R and r (R > r) have positive charge of equal surface density σ. The electrical potential at their common centre is
(a) \( \frac{σ}{ε_0} \)
(b) \( \frac{σ(R - r)}{ε_0} \)
(c) \( \frac{σ(R + r)}{ε_0} \)
(d) \( \frac{σ(R + r)}{ε_0 r} \)

Answer: (c) \( \frac{σ(R + r)}{ε_0} \)

1. Potential at the centre due to a charged shell = \( \frac{kQ}{radius} \) = \( \frac{1}{4π ε_0} \) × \( \frac{σ 4π R^{2}}{R} \) = \( \frac{σ R}{ε_0} \).
2. Similarly, the inner shell contributes \( \frac{σ r}{ε_0} \).
3. Total potential = \( \frac{σ(R + r)}{ε_0} \).
4. Hence, the correct option is (c).

Teacher's Note:
a) The potential inside a charged shell is constant and equals the potential on its surface.
b) Potentials are scalars and simply add.

3. Four charges, each equal to q, are placed at the corners of a square of side l. The electrical potential at the centre of the square is
(a) \( \frac{q}{π ε_0 l} \)
(b) \( \frac{1}{4π ε_0} \)\( \frac{4q}{\sqrt{2} l} \)
(c) \( \frac{1}{π ε_0} \)\( \frac{\sqrt{2} q}{l} \)
(d) \( \frac{2q}{π ε_0 l} \)

Answer: (c) \( \frac{1}{π ε_0} \)\( \frac{\sqrt{2} q}{l} \)

1. The distance of each corner from the centre = \( \frac{l}{\sqrt{2}} \).
2. V = 4 × \( \frac{1}{4π ε_0} \) × \( \frac{q}{l/\sqrt{2}} \).
3. V = \( \frac{4\sqrt{2} q}{4π ε_0 l} \) = \( \frac{1}{π ε_0} \) \( \frac{\sqrt{2} q}{l} \).
4. Hence, the correct option is (c).

Teacher's Note:
a) Half the diagonal, not half the side, is the distance from the centre to a corner.

4. An uncharged parallel plate condenser, having a dielectric constant K, is connected to a similar parallel plate condenser with air as medium in between plates and charged to a potential V. The two condensers are connected so that their common potential is V′. The dielectric constant K is
(a) \( \frac{V' - V}{V' + V} \)
(b) \( \frac{V' - V}{V'} \)
(c) \( \frac{V' - V}{V} \)
(d) \( \frac{V - V'}{V'} \)

Answer: (d) \( \frac{V - V'}{V'} \)

1. Let the capacitance of the air condenser be C, so that of the dielectric condenser is KC.
2. Charge is conserved: C V = (C + KC) V′.
3. 1 + K = \( \frac{V}{V'} \), so K = \( \frac{V - V'}{V'} \).
4. Hence, the correct option is (d).

Teacher's Note:
a) Total charge is conserved when two capacitors share charge, while the potential becomes common.

5. The charge and potential difference across 4 μF capacitor in the given circuit is
(a) 1200 μC, 310 V
(b) 600 μC, 310 V
(c) 1200 μC, 150 V
(d) 600 μC, 150 V

[Figure: A 310 V battery (positive terminal on the top) connected in a loop with a 20 μF capacitor in the top wire, a 12 μF capacitor in the bottom wire, and two 4 μF capacitors connected in parallel between the top and bottom wires on the right side]

Answer: (d) 600 μC, 150 V

1. The two 4 μF capacitors are in parallel: 4 + 4 = 8 μF.
2. This is in series with 20 μF and 12 μF: \( \frac{1}{C} \) = \( \frac{1}{20} \) + \( \frac{1}{12} \) + \( \frac{1}{8} \) = \( \frac{31}{120} \), so C = \( \frac{120}{31} \) μF.
3. Total charge = \( \frac{120}{31} \) × 310 = 1200 μC, which is the charge on the 20 μF, the 12 μF and the 8 μF combination.
4. Potential difference across the 8 μF combination (and across each 4 μF) = \( \frac{1200}{8} \) = 150 V.
5. Charge on each 4 μF capacitor = 4 × 150 = 600 μC.
6. Hence, the correct option is (d).

Teacher's Note:
a) In a series chain every capacitor (or parallel group) has the same charge; the charge divides equally between two equal parallel capacitors.
b) The printed key (c) gives 1200 μC, which is the charge on the whole parallel group and not on a single 4 μF capacitor.

6. A system of three parallel plates, each of area A, are separated by distances d₁ and d₂. The space between these plates is filled with dielectrics of relative permittivity ε₁ and ε₂ as shown in the figure. The permittivity of free space is ε₀. The equivalent capacitance of the system is
(a) \( \frac{ε_1 ε_2 A}{ε_2 d_1 + ε_1 d_2} \)
(b) \( \frac{ε_1 ε_2 ε_0 A}{ε_1 d_1 + ε_2 d_2} \)
(c) \( \frac{ε_0 A}{ε_1 d_1 + ε_2 d_2} \)
(d) \( \frac{ε_0 A}{ε_1 d_2 + ε_2 d_1} \)

[Figure: Three parallel vertical plates numbered 1, 2 and 3. Between plates 1 and 2, separated by distance d₁, the dielectric has permittivity ε₁; between plates 2 and 3, separated by distance d₂, the dielectric has permittivity ε₂]

Answer: (a) \( \frac{ε_1 ε_2 A}{ε_2 d_1 + ε_1 d_2} \)

1. The system is two capacitors in series: C₁ = \( \frac{ε_1 A}{d_1} \) and C₂ = \( \frac{ε_2 A}{d_2} \) (with ε₁ and ε₂ as the permittivities of the media).
2. \( \frac{1}{C} \) = \( \frac{d_1}{ε_1 A} \) + \( \frac{d_2}{ε_2 A} \) = \( \frac{ε_2 d_1 + ε_1 d_2}{ε_1 ε_2 A} \).
3. C = \( \frac{ε_1 ε_2 A}{ε_2 d_1 + ε_1 d_2} \).
4. Hence, the correct option is (a).

Teacher's Note:
a) Two dielectric layers between a common middle plate behave as two capacitors in series.
b) Only option (a) has the correct arrangement of d₁ and d₂ in the denominator; in it ε₁ and ε₂ stand for the permittivities of the media.

7. Five capacitors are connected as shown. The equivalent capacitance between terminals A and B and charge on 5μF capacitor will be respectively
(a) 8 μF, 100 μC
(b) 4 μF, 50 μC
(c) 12 μF, 150 μC
(d) 16 μF, 200 μC

[Figure: A chain between terminals B and A containing 12 μF, 10 μF, 5 μF, 9 μF and 8 μF capacitors in this order. A wire joins the point between 12 μF and 10 μF to the point between 5 μF and 9 μF; another wire joins the point between 10 μF and 5 μF to the point between 9 μF and 8 μF. A 60 V battery is connected between A (positive) and B]

Answer: (b) 4 μF, 50 μC

1. Name the junctions along the chain from B: a (after 12 μF), b (after 10 μF), c (after 5 μF), d (after 9 μF). The top wire joins a to c and the lower wire joins b to d.
2. So the 10 μF, 5 μF and 9 μF capacitors are all connected between the same two points (a-c and b-d): they are in parallel, giving 10 + 5 + 9 = 24 μF.
3. Equivalent capacitance: \( \frac{1}{C} \) = \( \frac{1}{12} \) + \( \frac{1}{24} \) + \( \frac{1}{8} \) = \( \frac{1}{4} \), so C = 4 μF.
4. Total charge = 4 × 60 = 240 μC, so the voltage across the 24 μF group = \( \frac{240}{24} \) = 10 V.
5. Charge on the 5 μF capacitor = 5 × 10 = 50 μC.
6. Hence, the correct option is (b).

Teacher's Note:
a) Redraw the circuit by joining the points connected by plain wires before combining capacitors.
b) Charge on one member of a parallel group is its capacitance times the group voltage.

8. A parallel plate capacitor of plate area A and separation between the plates d is charged to a potential difference V by a battery. The battery is then disconnected and a slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between plates completely. The work done to insert the slab is
(a) \( \frac{ε_0 A V^{2}}{2d} \)(\( \frac{1}{K} \) - 1)
(b) \( \frac{ε_0 A V^{2} (K - 1)}{2d} \)
(c) \( \frac{ε_0 A V^{2}}{d(K - 1)} \)
(d) \( \frac{ε_0 A V^{2} K}{d} \)

Answer: (a) \( \frac{ε_0 A V^{2}}{2d} \)(\( \frac{1}{K} \) - 1)

1. Initial energy U_i = \( \frac{1}{2} \) C V² = \( \frac{ε_0 A V^{2}}{2d} \).
2. With the battery disconnected, the charge is constant, so the final energy U_f = \( \frac{Q^{2}}{2KC} \) = \( \frac{U_i}{K} \).
3. Work done = U_f - U_i = \( \frac{ε_0 A V^{2}}{2d} \)(\( \frac{1}{K} \) - 1).
4. This is negative because the slab is pulled into the capacitor by the field.
5. Hence, the correct option is (a).

Teacher's Note:
a) At constant charge the energy of a capacitor decreases when a dielectric is inserted.
b) The negative sign means the capacitor does work on the slab.

9. Three charges +q, +q and Q are palced at the vertices of a right-angled isosceles triangle as shown in the figure. The net electrostatic potential energy of the configuration is zero if Q is equal to
(a) \( \frac{-q}{\sqrt{2} + 1} \)
(b) \( \frac{-\sqrt{2} q}{\sqrt{2} + 1} \)
(c) \( \frac{q}{2 + \sqrt{2}} \)
(d) -2q

[Figure: A right-angled isosceles triangle with the right angle (90°) at the top vertex carrying charge +q. The two equal sides, each of length a, run down to the left vertex carrying charge +q and to the right vertex carrying charge Q]

Answer: (b) \( \frac{-\sqrt{2} q}{\sqrt{2} + 1} \)

1. The legs have length a, and the hypotenuse between +q (at the left) and Q has length a√2.
2. U = \( \frac{1}{4π ε_0} \)[\( \frac{q^{2}}{a} \) + \( \frac{q Q}{a} \) + \( \frac{q Q}{\sqrt{2} a} \)].
3. For U = 0: q + Q(1 + \( \frac{1}{\sqrt{2}} \)) = 0, so Q = \( \frac{-q}{1 + 1/\sqrt{2}} \).
4. Q = \( \frac{-\sqrt{2} q}{\sqrt{2} + 1} \).
5. Hence, the correct option is (b).

Teacher's Note:
a) Potential energy of a system is the sum over all pairs; count each pair once.
b) Use the distance a√2 for the pair along the hypotenuse.

10. Two equal negative charges -q are fixed at points with coordinates (0, a) and (0, -a) on y-axis. A positive charge Q is released from rest at the point with coordinates (2a, 0) on x-axis. The charge Q will
(a) execute simple harmonic motion about the origin
(b) move to origin and remain at rest
(c) move to infinity
(d) execute oscillatory motion but not simple harmonic motion

Answer: (d) execute oscillatory motion but not simple harmonic motion

1. The positive charge Q is attracted towards the origin by both negative charges, and the net force is along the x-axis towards the origin.
2. It crosses the origin with maximum speed and moves to the other side, where the force again acts towards the origin, so it oscillates along the x-axis.
3. The net force F = \( \frac{2 k q Q x}{(x^{2} + a^{2})^{3/2}} \) is not proportional to x for large displacements such as x = 2a, so the motion is not simple harmonic.
4. Hence, the correct option is (d).

Teacher's Note:
a) Motion is simple harmonic only when the restoring force is proportional to displacement; here this holds only for x much smaller than a.

11. A capacitor C₁ is charged to a potential difference V. The charging battery is then removed and the capacitor is connected to an uncharged capacitor C₂, If U is the initial energy in capacitor C₁, the final energy stored in the combination is
(a) U(\( \frac{C_1}{C_1 + C_2} \))
(b) U(\( \frac{C_2}{C_1 + C_2} \))
(c) \( \frac{U C_2 C_1}{C_1 + C_2} \)
(d) \( \frac{U}{C_1 + C_2} \)

Answer: (a) U(\( \frac{C_1}{C_1 + C_2} \))

1. Initial energy U = \( \frac{1}{2} \) C₁ V² and the charge Q = C₁V.
2. The common potential = \( \frac{C_1 V}{C_1 + C_2} \).
3. Final energy = \( \frac{1}{2} \) (C₁ + C₂) \( \frac{C_1^{2} V^{2}}{(C_1 + C_2)^{2}} \) = \( \frac{C_1}{C_1 + C_2} \) × \( \frac{1}{2} \) C₁V² = U \( \frac{C_1}{C_1 + C_2} \).
4. Hence, the correct option is (a).

Teacher's Note:
a) Energy is lost (as heat and radiation) when a charged capacitor shares charge with an uncharged one.

12. The charge on the capacitor of 5 μF in the circuit given in the figure is
(a) 7 μC
(b) 9 μC
(c) 12.6 μC
(d) 21 μC

[Figure: A 6 V battery with a 4 μF capacitor connected directly across it in one branch. In another branch across the same battery, a 3 μF capacitor is in series with a parallel pair of a 2 μF capacitor and a 5 μF capacitor]

Answer: (b) 9 μC

1. The 2 μF and 5 μF capacitors are in parallel: 2 + 5 = 7 μF. This is in series with the 3 μF capacitor.
2. Equivalent of this branch = \( \frac{3 \times 7}{3 + 7} \) = 2.1 μF. The 4 μF capacitor is in a separate parallel branch across the 6 V battery.
3. Charge on the 3 μF capacitor and on the 7 μF group = 2.1 × 6 = 12.6 μC.
4. Potential difference across the 7 μF group = \( \frac{12.6}{7} \) = 1.8 V.
5. Charge on the 5 μF capacitor = 5 × 1.8 = 9 μC.
6. Hence, the correct option is (b).

Teacher's Note:
a) The 4 μF capacitor is in parallel with the other branch and has no effect on the charge on the 5 μF capacitor.
b) Charge divides between parallel capacitors in proportion to their capacitances: 12.6 × 5/7 = 9 μC.

13. A capacitor of 1 μF withstands a maximum voltage of 6 KV while another capacitor of 2 μF withstands a maximum voltage of 4 KV. If the two capacitors are connected in series, the system will withstand a maximum voltage of
(a) 2 KV
(b) 4 KV
(c) 6 KV
(d) 9 KV

Answer: (d) 9 KV

1. Maximum charge on the first capacitor = 1 μF × 6 kV = 6000 μC and on the second = 2 μF × 4 kV = 8000 μC.
2. In series both carry the same charge, so the maximum safe charge = 6000 μC.
3. Voltage across the 1 μF = 6 kV and across the 2 μF = \( \frac{6000}{2} \) = 3 kV.
4. Maximum voltage of the system = 6 + 3 = 9 kV.
5. Hence, the correct option is (d).

Teacher's Note:
a) In series the charge is common, so the capacitor that reaches its limit first decides the maximum charge.

14. Two metallic spheres of radii 20 cm and 10 cm are each given an equal charge of + 150 μC. If these are connected by a conducting wire, their common potential is
(a) 4.5 × 10⁶ V
(b) 9 × 10⁶ V
(c) 1.35 × 10⁷ V
(d) 1.8 × 10⁷ V

Answer: (b) 9 × 10⁶ V

1. Total charge = 150 + 150 = 300 μC = 3 × 10⁻⁴ C.
2. Common potential V = \( \frac{k (Q_1 + Q_2)}{R_1 + R_2} \) = \( \frac{9 \times 10^{9} \times 3 \times 10^{-4}}{0.2 + 0.1} \).
3. V = 9 × 10⁶ V.
4. Hence, the correct option is (b).

Teacher's Note:
a) After connection, the total charge is shared so that both spheres reach the same potential.
b) The common potential is the total charge divided by the sum of the capacitances (radii).

15. A parallel plate capacitor of 1.77 μF is to be designed using a dielectric material of dielelctric constant 200 and of breakdown strength 3 × 10⁶ V/m. In order to make a capacitor which can withstand a potential difference of 20 V across the plates, the separation between plates, l and area of plates A, will be respectively
(a) l = 10⁻⁶ m, A = 10⁻³m²
(b) l = 10⁻⁵ m, A = 10⁻²m²
(c) l = 10⁻⁴m, A = 10⁻⁴m²
(d) l = 10⁻⁴m, A = 10⁻⁶m²

Answer: (b) l = 10⁻⁵ m, A = 10⁻²m²

1. For the capacitor not to break down, \( \frac{20}{l} \) must not exceed 3 × 10⁶ V/m, so l is not less than 6.7 × 10⁻⁶ m.
2. This rules out l = 10⁻⁶ m, because the field would be 2 × 10⁷ V/m.
3. Capacitance C = \( \frac{K ε_0 A}{l} \) = 200 × 8.85 × 10⁻¹² × \( \frac{A}{l} \) = 1.77 × 10⁻⁹ \( \frac{A}{l} \).
4. For C = 1.77 × 10⁻⁶ F, \( \frac{A}{l} \) = 10³ m. Of the remaining options, l = 10⁻⁵ m with A = 10⁻² m² satisfies this ratio.
5. Field = \( \frac{20}{10^{-5}} \) = 2 × 10⁶ V/m, which is below the breakdown strength.
6. Hence, the correct option is (b).

Teacher's Note:
a) Two conditions must both hold: the required capacitance and a field below the breakdown strength.
b) The printed key (a) fits the capacitance, but its field of 2 × 10⁷ V/m exceeds the breakdown strength, so it is not acceptable.

16. Two slabs of the same dimensions having dielectric constants K₁ and K₂ completely fill the space between plates of a parallel plate capacitor as shown. If C is original capacitance of capacitor and A is original area, the new capacitance is
(a) \( \frac{2K_1 K_2}{K_1 + K_2} \) C
(b) \( \frac{K_1 + K_2}{2} \) C
(c) (K₁ + K₂) C
(d) \( \frac{K_1 K_2}{K_1 + K_2} \) C

[Figure: A parallel plate capacitor of plate separation d (shown as the horizontal width). The space is divided into an upper half of area A/2 filled with dielectric K₁ and a lower half of area A/2 filled with dielectric K₂, both extending across the whole separation d]

Answer: (b) \( \frac{K_1 + K_2}{2} \) C

1. Each slab fills the full separation d but covers half the area, \( \frac{A}{2} \). So the two capacitors are in parallel.
2. C₁ = \( \frac{K_1 ε_0 A/2}{d} \) = \( \frac{K_1}{2} \) C and C₂ = \( \frac{K_2}{2} \) C.
3. New capacitance = C₁ + C₂ = \( \frac{K_1 + K_2}{2} \) C.
4. Hence, the correct option is (b).

Teacher's Note:
a) Slabs side by side (each spanning the whole gap) are in parallel; slabs stacked one after another along the gap are in series.

17. Two slabs of the same dimensions having dielectric constants K₁ and K₂ completely fill the space between plates of a parallel plate capacitor as shown in the figure. If C is original capacitance of the capacitor, the new capacitance is
(a) \( \frac{K_1 + K_2}{2} \) C
(b) \( \frac{2K_1 K_2}{K_1 + K_2} \) C
(c) (K₁ + K₂) C
(d) \( \frac{K_1 K_2}{K_1 + K_2} \) C

[Figure: A parallel plate capacitor with the plates at the top and bottom. The gap between them is filled by two slabs of equal width d/2 placed side by side, the left slab with dielectric K₁ and the right slab with dielectric K₂, with total width d marked below]

Answer: (b) \( \frac{2K_1 K_2}{K_1 + K_2} \) C

1. The slabs are placed one after the other across the gap, each of thickness \( \frac{d}{2} \), so the two capacitors are in series.
2. C₁ = \( \frac{K_1 ε_0 A}{d/2} \) = 2K₁C and C₂ = 2K₂C.
3. New capacitance = \( \frac{C_1 C_2}{C_1 + C_2} \) = \( \frac{4K_1 K_2 C^{2}}{2(K_1 + K_2) C} \) = \( \frac{2K_1 K_2}{K_1 + K_2} \) C.
4. Hence, the correct option is (b).

Teacher's Note:
a) Layers stacked along the field direction share the same flux density and are in series.

18. A solid sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell is V. If the shell is now given a charge of -3Q, the new potential difference between the two surfaces is
(a) V
(b) 2V
(c) 4V
(d) -2V

Answer: (a) V

1. A charge on the shell adds the same constant potential at every point inside and on the shell.
2. The potential difference between the sphere and the shell depends only on the charge Q of the sphere and the radii.
3. So the potential difference stays V after the shell is given a charge of -3Q.
4. Hence, the correct option is (a).

Teacher's Note:
a) The potential difference between two concentric conductors depends only on the charge of the inner one.
b) The charge on the shell shifts both potentials equally and cancels in the difference.

19. Seven capaciotrs, each of capacitance 2 μF are to be connected in a configuration to obtain an effective capacitance of (10/11) μF. Which of the combination(s), as shown in the figure, will achieve the desired result.
(a) 5 capacitors in parallel, in series with 2 capacitors in series
(b) 4 capacitors in parallel, in series with 3 capacitors in series
(c) 3 capacitors in parallel, in series with 4 capacitors in series
(d) 2 capacitors in parallel, in series with 5 capacitors in series

[Figure: Four arrangements of seven 2 μF capacitors between two end points. (a) Five capacitors connected in parallel between two junctions, followed by two capacitors in series. (b) Four capacitors in parallel, followed by three capacitors in series. (c) Three capacitors in parallel, followed by four capacitors in series. (d) Two capacitors in parallel, followed by five capacitors in series]

Answer: (a) 5 capacitors in parallel, in series with 2 capacitors in series

1. Option (a): five capacitors in parallel give 10 μF. Two capacitors in series give 1 μF. Together in series: \( \frac{10 \times 1}{10 + 1} \) = \( \frac{10}{11} \) μF.
2. Option (b): 8 μF in series with \( \frac{2}{3} \) μF gives \( \frac{8}{13} \) μF, which is not required.
3. Option (c): 6 μF in series with 0.5 μF gives \( \frac{6}{13} \) μF, which is not required.
4. Option (d): 4 μF in series with 0.4 μF gives \( \frac{4}{11} \) μF, which is not required.
5. Hence, the correct option is (a).

Teacher's Note:
a) Count the capacitors in each figure: the total must be seven.
b) Parallel groups add capacitance and series groups add the reciprocals.

20. A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its centre
(a) increases as r increases for r > R
(b) decreases as r increases for 0 < r < ∞
(c) increases as r increases for R < r < ∞
(d) is discontinuous at r = R

Answer: (a) increases as r increases for r > R

1. Inside a uniformly charged non-conducting sphere (r less than R), E = \( \frac{ρ r}{3ε_0} \), so E increases linearly with r.
2. Outside (r greater than R), E = \( \frac{kQ}{r^{2}} \), which decreases as r increases.
3. At r = R the two expressions are equal, so E is continuous.
4. Option (a) is the statement intended for r less than R; options (b), (c) and (d) are wrong for the reasons above, so the correct option is (a).

Teacher's Note:
a) Option (a) as printed says "for r > R", which is a misprint for "r < R"; the field outside the sphere actually decreases.
b) A uniformly charged non-conducting sphere has a continuous field, while a charged conducting shell has a field that jumps at its surface.

21. A charge +q is fixed at each of the points x = x₀, x = 3x₀, x = 5x₀ ........... up to infinity on the x-axis and a charge -q is fixed at each of the points x = 2x₀, x = 4x₀ and x = 6x₀ ........... where x is a positive constant. Take the electric potential at a point due to a charge +Q at a distance r from it to be \( \frac{Q}{4π ε_0 r} \). Then the potential at the origin due to the above system of charges is .........
(a) 0
(b) \( \frac{q}{8π ε_0 \ln 2} \)
(c) ∞
(d) \( \frac{q \ln 2}{4π ε_0 x_0} \)

Answer: (d) \( \frac{q \ln 2}{4π ε_0 x_0} \)

1. Potential at the origin V = \( \frac{q}{4π ε_0 x_0} \)[1 - \( \frac{1}{2} \) + \( \frac{1}{3} \) - \( \frac{1}{4} \) + ...].
2. The series 1 - \( \frac{1}{2} \) + \( \frac{1}{3} \) - ... = ln 2.
3. V = \( \frac{q \ln 2}{4π ε_0 x_0} \).
4. Hence, the correct option is (d).

Teacher's Note:
a) The alternating series is the expansion of ln(1 + x) at x = 1.

22. Two conducting balls of same mass M and radius r are suspended from insulating threads of length L each and are given charge Q each as shown in the figure. If the angular displacement θ from the vertical is small, the distance d between the balls is
(a) (\( \frac{Q^{2} L}{4π ε_0 M g} \))^1/3
(b) (\( \frac{Q^{2} L}{4π ε_0 M g} \))^3/2
(c) (\( \frac{Q^{2} L}{2π ε_0 M g} \))^1/3
(d) (\( \frac{Q^{2} L}{2π ε_0 M g} \))^1/2

[Figure: Two balls of mass M, each carrying charge Q, hang from a common point by insulating threads of length L. Each thread makes an angle θ with the vertical dotted line, and the balls are a distance d apart on a horizontal line]

Answer: (c) (\( \frac{Q^{2} L}{2π ε_0 M g} \))^1/3

1. Each ball is in equilibrium under gravity Mg, the tension T and the electrostatic force F = \( \frac{Q^{2}}{4π ε_0 d^{2}} \).
2. tan θ = \( \frac{F}{Mg} \), and for small θ, tan θ ≈ \( \frac{d/2}{L} \).
3. \( \frac{Q^{2}}{4π ε_0 d^{2} M g} \) = \( \frac{d}{2L} \), so d³ = \( \frac{Q^{2} L}{2π ε_0 M g} \).
4. d = (\( \frac{Q^{2} L}{2π ε_0 M g} \))^1/3, so the correct option is (c).

Teacher's Note:
a) The horizontal distance of each ball from the vertical is d/2, and tan θ is replaced by sin θ for small angles.

23. A particle of charge Q and mass M is placed at rest in a uniform electric field E as shown and released. The kinetic energy imparted to the particle, neglecting gravitation force, after moving a distance x, is
(a) QEx
(b) \( \frac{QE x}{2} \)
(c) QE²x
(d) \( \frac{QE x}{4} \)

[Figure: A particle of charge Q between a positively charged plate on the left and a negatively charged plate on the right, with the uniform field E between them]

Answer: (a) QEx

1. The force on the particle is QE and it moves a distance x along the force.
2. Work done = force × distance = QEx.
3. By the work-energy theorem, the kinetic energy = QEx.
4. Hence, the correct option is (a).

Teacher's Note:
a) The kinetic energy gained equals the work done by the field, which is the charge times the potential difference.

24. It is required to proudce an electric field of 10⁻³ N/C at the surface of a spherical conductor of radius 0.1 m. The number of electrons which should be added to the spherical conductor is
(a) 1.4 × 10⁴
(b) 6.9 × 10³
(c) 1.4 × 10⁵
(d) 6.9 × 10⁴

Answer: (b) 6.9 × 10³

1. E = \( \frac{kQ}{r^{2}} \), so Q = \( \frac{E r^{2}}{k} \) = \( \frac{10^{-3} \times 0.01}{9 \times 10^{9}} \) = 1.1 × 10⁻¹⁵ C.
2. Number of electrons n = \( \frac{Q}{e} \) = \( \frac{1.1 \times 10^{-15}}{1.6 \times 10^{-19}} \).
3. n ≈ 6.9 × 10³.
4. Hence, the correct option is (b).

Teacher's Note:
a) Use the radius of the sphere as the distance for the field at its surface.

25. A potential difference V is applied to n identical capacitors connected in parallel. These capacitors are now connected in series without disturbing their charges. The potential difference of the capacitors connected in series is
(a) \( \frac{V}{n} \)
(b) nV
(c) n²V
(d) (n - 1) V

Answer: (b) nV

1. In parallel, each capacitor has charge q = CV.
2. When reconnected in series (positive plate of one to the negative plate of the next) without disturbing the charges, the charge on each remains q and the potentials add.
3. Total potential difference = n × \( \frac{q}{C} \) = nV.
4. Hence, the correct option is (b).

Teacher's Note:
a) Reconnecting capacitors in series with the charges kept the same adds up their individual voltages.

26. A charge q is fixed at each of the two opposite corners of a square of side l. A charge Q is placed at each of the other two corners. If the resultant electrical force on q is zero, then q is given by
(a) √2 Q
(b) -2√2 Q
(c) √2 Q²
(d) -2√2 Q²

Answer: (b) -2√2 Q

1. On one charge q, the other q at distance √2 l repels it with force \( \frac{k q^{2}}{2 l^{2}} \) along the diagonal.
2. Each charge Q at distance l exerts a force \( \frac{k q Q}{l^{2}} \); the two forces are at 90°, and their resultant = \( \frac{\sqrt{2} k q Q}{l^{2}} \) along the diagonal.
3. For zero net force, the forces along the diagonal must cancel: \( \frac{q^{2}}{2} \) = -√2 qQ, so q = -2√2 Q.
4. Hence, the correct option is (b).

Teacher's Note:
a) Resolve the forces along the diagonal; the sign of q must be opposite to that of Q for the forces to cancel.

27. A parallel plate capacitor with air as medium has a capacitance of 14 μF. The capacitor is filled with three media of dielectric constant K₁ = 2, K₂ = 3 and K₃ = 4. The capacitance of the system will be
(a) 19 μF
(b) 21 μF
(c) 25 μF
(d) 38 μF

[Figure: A parallel plate capacitor of plate separation d. The left half of the area is filled with dielectric K₁ over the full separation d. The right half of the area is split into an upper layer of thickness d/2 filled with K₂ and a lower layer of thickness d/2 filled with K₃]

Answer: (d) 38 μF

1. The left half (area \( \frac{A}{2} \), full gap d, dielectric K₁ = 2) has capacitance C₁ = 2 × \( \frac{14}{2} \) = 14 μF.
2. The right half has area \( \frac{A}{2} \) and two layers, each of thickness \( \frac{d}{2} \), with K₂ = 3 and K₃ = 4 in series. Each layer on its own has capacitance 2K × \( \frac{14}{2} \) = 14K μF, so the layers are 42 μF and 56 μF.
3. In series: \( \frac{42 \times 56}{42 + 56} \) = 24 μF.
4. Total capacitance = 14 + 24 = 38 μF.
5. Hence, the correct option is (d).

Teacher's Note:
a) Treat the left half and the right half as parallel capacitors, and the two layers in the right half as a series pair.

28. The plates of a parallel plate capacitor are charged upto 100 V. A 2 mm thick plate is inserted between the plates. In order to maintain the same potential difference, the distance between the plates is increased by 1.6 mm. The dielectric constant of the plate is
(a) 1.25
(b) 2.5
(c) 4
(d) 5

Answer: (d) 5

1. With a slab of thickness t and dielectric constant K inserted in a gap d, the effective gap is d - t + \( \frac{t}{K} \).
2. For the same potential difference at the same charge, the effective gap must remain d, so (d + 1.6) - 2 + \( \frac{2}{K} \) = d.
3. \( \frac{2}{K} \) = 2 - 1.6 = 0.4, so K = 5.
4. Hence, the correct option is (d).

Teacher's Note:
a) A dielectric slab is equivalent to removing a thickness t and adding t/K of air gap.

29. An electric dipole of moment p⃗ is placed in an equilibrium position in a uniform electric field of intensity E. The potential energy of the dipole when it is rotated through an angle θ from the mean position is
(a) pEsinθ
(b) pEcosθ
(c) pE(1 - sinθ)
(d) pE(1 - cosθ)

Answer: (d) pE(1 - cosθ)

1. Potential energy of a dipole at angle φ to the field is U = -pE cos φ.
2. At equilibrium (φ = 0) U = -pE; after rotating through θ, U = -pE cos θ.
3. The potential energy measured from the equilibrium position = -pE cos θ - (-pE) = pE(1 - cos θ).
4. Hence, the correct option is (d).

Teacher's Note:
a) The work done in rotating the dipole from the stable equilibrium position to angle θ is pE(1 - cos θ).

30. An electric dipole consisting of two opposite charges of 2 μC separated by a distance of 3 cm is placed in an electric field of 10⁵ NC⁻¹. The maximum torque on the dipole will be
(a) 2 × 10⁵ N-m
(b) 6 × 10⁻³ N-m
(c) 1.2 × 10⁻² N-m
(d) 1.2 N-m

Answer: (b) 6 × 10⁻³ N-m

1. Dipole moment p = q × 2a = 2 × 10⁻⁶ × 0.03 = 6 × 10⁻⁸ C m.
2. Maximum torque = pE (when the dipole is perpendicular to the field) = 6 × 10⁻⁸ × 10⁵.
3. Maximum torque = 6 × 10⁻³ N-m.
4. Hence, the correct option is (b).

Teacher's Note:
a) Convert 3 cm to 0.03 m and μC to 10⁻⁶ C.

31. Two points P₁ and P₂ are symmetrical along the axis of a small electric dipole. If E₁ and E₂ are electric intensities and V₁ and V₂ are electric potentials at these points, then
(a) E₁ = E₂ & V₁ = V₂
(b) E₁ = - E₂ & V₁ = - V₂
(c) E₁ = E₂ & V₁ = - V₂
(d) E₁ = - E₂ & V₁ = V₂

Answer: (c) E₁ = E₂ & V₁ = - V₂

1. On the axis, the field at a point at distance r is E = \( \frac{2kp}{r^{3}} \) directed along the dipole moment, on both sides of the dipole. So E₁ = E₂ in magnitude and direction.
2. The potential on the axis is V = ± \( \frac{kp}{r^{2}} \), positive on the side of the positive charge and negative on the other side.
3. So V₁ = -V₂.
4. Hence, the correct option is (c).

Teacher's Note:
a) The axial field points the same way on both sides of the dipole, while the potential changes sign.

32. Identical charges (- q each) are placed at each corner of a cube of side b. The electrical P.E. of charge (+ q), placed at its centre, will be
(a) \( \frac{-4\sqrt{2} q^{2}}{π ε_0 b} \)
(b) \( \frac{-8\sqrt{2} q^{2}}{π ε_0 b} \)
(c) \( \frac{-4 q^{2}}{\sqrt{3} π ε_0 b} \)
(d) \( \frac{-8\sqrt{2} q^{2}}{4π ε_0 b} \)

Answer: (c) \( \frac{-4 q^{2}}{\sqrt{3} π ε_0 b} \)

1. The distance from the centre of the cube to a corner = \( \frac{\sqrt{3} b}{2} \).
2. The potential energy of +q with the eight charges = 8 × \( \frac{1}{4π ε_0} \) × \( \frac{(q)(-q)}{\sqrt{3} b/2} \).
3. U = -\( \frac{16 q^{2}}{4\sqrt{3} π ε_0 b} \) = \( \frac{-4 q^{2}}{\sqrt{3} π ε_0 b} \).
4. Hence, the correct option is (c).

Teacher's Note:
a) Only the interaction of +q with the corner charges is asked, not the energy of the corner charges among themselves.
b) The body diagonal of a cube is √3 b, so half of it is the centre-to-corner distance.

33. Two infinitely long parallel wires having linear charge densities λ₁ and λ₂, respectively, are placed at a distance of R metres. The force per unit length on either wire will be (K = 1/4 π ∈₀)
(a) K \( \frac{2 λ_1 λ_2}{R^{2}} \)
(b) K \( \frac{2 λ_1 λ_2}{R} \)
(c) K \( \frac{λ_1 λ_2}{R^{2}} \)
(d) K \( \frac{λ_1 λ_2}{R} \)

Answer: (b) K \( \frac{2 λ_1 λ_2}{R} \)

1. Field of the first wire at the distance R, E = \( \frac{2Kλ_1}{R} \).
2. Force per unit length on the second wire = λ₂ E = \( \frac{2Kλ_1 λ_2}{R} \).
3. Hence, the correct option is (b).

Teacher's Note:
a) For an infinite line charge, the field falls as 1/R, so the force per unit length falls as 1/R and not 1/R².

34. A charge Q is distributed over two concentric hollow spheres of radii R and 2R such that the surface densities are equal. The potential at the common centre is
(a) \( \frac{2}{3} \) · \( \frac{Q}{4π ε_0 R} \)
(b) \( \frac{3}{4} \) · \( \frac{Q}{4π ε_0 R} \)
(c) \( \frac{2}{5} \) · \( \frac{Q}{4π ε_0 R} \)
(d) \( \frac{3}{5} \) · \( \frac{Q}{4π ε_0 R} \)

Answer: (d) \( \frac{3}{5} \) · \( \frac{Q}{4π ε_0 R} \)

1. Equal surface densities give charges in the ratio of the areas: Q₁ : Q₂ = R² : 4R² = 1 : 4, so Q₁ = \( \frac{Q}{5} \) and Q₂ = \( \frac{4Q}{5} \).
2. Potential at the centre = \( \frac{1}{4π ε_0} \)[\( \frac{Q/5}{R} \) + \( \frac{4Q/5}{2R} \)] = \( \frac{1}{4π ε_0} \) \( \frac{Q}{R} \)(\( \frac{1}{5} \) + \( \frac{2}{5} \)).
3. V = \( \frac{3}{5} \) · \( \frac{Q}{4π ε_0 R} \).
4. Hence, the correct option is (d).

Teacher's Note:
a) Equal surface density means charge is proportional to the surface area, that is, to the square of the radius.

35. Two concentric hollow metal spheres have radii R₁ and R₂. The outer sphere of radius R₂ is given a positive charge q and the inner is earthed. The charge on the inner sphere is
(a) zero
(b) -q
(c) -R₁q(R₁+R₂)
(d) \( \frac{-R_1 q}{R_2} \)

[Figure: Two concentric hollow metal spheres: a smaller inner sphere of radius R₁ joined by a wire to the earth symbol, and a larger outer sphere of radius R₂ carrying a charge +q]

Answer: (d) \( \frac{-R_1 q}{R_2} \)

1. Let the charge on the inner sphere be q₁. The inner sphere is earthed, so its potential is zero.
2. Potential of the inner sphere = \( \frac{k q_1}{R_1} \) + \( \frac{k q}{R_2} \) = 0. Here \( \frac{k q}{R_2} \) is the contribution of the outer sphere of total charge q; the charge induced on the outer sphere by q₁ adds up to zero in the potential.
3. \( \frac{q_1}{R_1} \) = -\( \frac{q}{R_2} \), so q₁ = \( \frac{-R_1 q}{R_2} \).
4. Hence, the correct option is (d).

Teacher's Note:
a) An earthed conductor is at zero potential, which fixes the induced charge on it.
b) The result lies between 0 and -q, since R₁ is less than R₂.