JEE Physics Electrostatic Potential and Capacitance MCQs Set 23

Physics Objective Questions and Answers: Electrostatic Potential and Capacitance

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Multiple Choice Questions

Question 1: The earth has volume V and surface area A. Its capacitance would be
(a) \( 4\pi\epsilon_0A/V \)
(b) \( 4\pi\epsilon_0V/A \)
(c) \( 12\pi\epsilon_0V/A \)
(d) \( 12\pi\epsilon_0A/V \)
Show Answer & Explanation

Answer: (c) \( 12\pi\epsilon_0V/A \)

Explanation:
1. \( C = 4\pi\epsilon_0R \).
2. \( V/A = (\tfrac43\pi R^3)/(4\pi R^2) = R/3 \), so \( R = 3V/A \).
3. \( C = 12\pi\epsilon_0V/A \).

Teacher's Note:
1. Capacitance of a sphere is proportional to its radius.
2. Express R using V/A.

Question 2: Two spherical conductors A and B of radii a and b (b > a) are placed in air concentrically. B is given a charge +Q coulomb and A is grounded. The equivalent capacitance of these is
(a) \( 4\pi\epsilon_0\dfrac{ab}{b-a} \)
(b) \( 4\pi\epsilon_0(a+b) \)
(c) \( 4\pi\epsilon_0b \)
(d) \( 4\pi\epsilon_0\dfrac{b^2}{b-a} \)
Show Answer & Explanation

Answer: (d) \( 4\pi\epsilon_0\dfrac{b^2}{b-a} \)

Explanation:
1. Between A and B there is a spherical capacitor \( 4\pi\epsilon_0ab/(b-a) \).
2. The outer surface of B forms a capacitor with the earth, \( 4\pi\epsilon_0b \), in parallel with it.
3. Total \( = 4\pi\epsilon_0\left[\dfrac{ab}{b-a}+b\right] = 4\pi\epsilon_0\dfrac{b^2}{b-a} \).

Teacher's Note:
1. The inner sphere is earthed, so both surfaces of B store charge.
2. The two capacitances are in parallel.

Question 3: A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is decreased using an insulating handle. As a result the potential difference between the plates
(a) decreases
(b) does not change
(c) becomes zero
(d) increases
Show Answer & Explanation

Answer: (a) decreases

Explanation:
1. With the battery removed, Q is constant.
2. Reducing d raises C, and \( V = Q/C \) decreases.

Teacher's Note:
1. Constant charge means V varies inversely with C.

Question 4: Two capacitors of capacity 2C and C/2 are connected in parallel to a V-volt battery. The work done by the battery in charging fully both the capacitors is
(a) \( \tfrac12CV^2 \)
(b) \( 2CV^2 \)
(c) \( \tfrac52CV^2 \)
(d) \( \tfrac32CV^2 \)
Show Answer & Explanation

Answer: (c) \( \tfrac52CV^2 \)

Explanation:
1. Parallel capacitance \( = 2C + C/2 = 5C/2 \).
2. Work done by the battery \( = C_{eq}V^2 = \tfrac52CV^2 \).

Teacher's Note:
1. The battery supplies QV, not QV/2.

Question 5: A series combination of n₁ capacitors, each of value C₁, is charged by a source of potential difference 8V. When another parallel combination of n₂ capacitors, each of value C₂, is charged by a source of potential difference 2V, it has the same (total) energy stored in it, as the first combination has. The value of C₂ in terms of C₁ is then
(a) \( 16C_1/(n_1n_2) \)
(b) \( 2C_1/(n_1n_2) \)
(c) \( 16\dfrac{n_2}{n_1}C_1 \)
(d) \( 2\dfrac{n_2}{n_1}C_1 \)
Show Answer & Explanation

Answer: (a) \( 16C_1/(n_1n_2) \)

Explanation:
1. First: \( C = C_1/n_1 \), \( U = \tfrac12(C_1/n_1)(8V)^2 = 32C_1V^2/n_1 \).
2. Second: \( C = n_2C_2 \), \( U = \tfrac12n_2C_2(2V)^2 = 2n_2C_2V^2 \).
3. Equating, \( C_2 = 16C_1/(n_1n_2) \).

Teacher's Note:
1. Series gives C/n, parallel gives nC.
2. Equate the two energies.

Question 6: Two parallel metal plates having charges +Q and −Q face each other with a certain separation between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will
(a) increase
(b) decrease
(c) remain same
(d) become zero
Show Answer & Explanation

Answer: (b) decrease

Explanation:
1. The charge stays constant, but the field in a dielectric is \( E = \sigma/(K\epsilon_0) \).
2. K > 1 for kerosene, so the field decreases.

Teacher's Note:
1. Dielectric polarisation reduces the net field.

Question 7: The energy stored between the plates of a parallel plate capacitor of area A, separation d, with electric field E between the plates is
(a) \( \tfrac12\epsilon_0E^2Ad \)
(b) \( \epsilon_0E^2Ad \)
(c) \( \tfrac12\epsilon_0E^2A/d \)
(d) \( \tfrac12\epsilon_0E^2d/A \)
Show Answer & Explanation

Answer: (a) \( \tfrac12\epsilon_0E^2Ad \)

Explanation:
1. Energy density \( = \tfrac12\epsilon_0E^2 \).
2. Volume between the plates \( = Ad \).
3. \( U = \tfrac12\epsilon_0E^2Ad \).

Teacher's Note:
1. Energy density times volume gives the total energy.

Question 8: A 300 pF capacitor is charged with a 100 V battery. After disconnecting the battery this capacitor is connected with another uncharged 300 pF capacitor. Then the energy loss is
(a) 25 × 10⁻⁸ J
(b) 50 × 10⁻⁸ J
(c) 75 × 10⁻⁸ J
(d) 100 × 10⁻⁸ J
Show Answer & Explanation

Answer: (c) 75 × 10⁻⁸ J

Explanation:
1. Initial energy \( = \tfrac12(300\times10^{-12})(100)^2 = 150\times10^{-8} \) J.
2. Identical capacitors share charge equally, so half of the energy is lost: 75 × 10⁻⁸ J.

Teacher's Note:
1. When equal capacitors share charge, half the energy is lost.

Question 9: A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?
(a) The energy stored in the capacitor decreases K times.
(b) The change in energy stored is \( \tfrac12CV^2(1/K - 1) \).
(c) The charge on the capacitor is not conserved.
(d) The potential difference between the plates decreases K times.
Show Answer & Explanation

Answer: (c) The charge on the capacitor is not conserved.

Explanation:
1. After disconnection the isolated plates keep their charge, so the statement that charge is not conserved is wrong.
2. With Q fixed, \( U = Q^2/2KC \) falls K times and V falls K times.

Teacher's Note:
1. Isolated plates cannot lose charge.

Question 10: A parallel plate air capacitor has capacity C and separation between the plates is d. A potential difference V is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is
(a) \( C^2V^2/2d^2 \)
(b) \( C^2V^2/2d \)
(c) \( CV^2/2d \)
(d) \( CV^2/d \)
Show Answer & Explanation

Answer: (c) \( CV^2/2d \)

Explanation:
1. \( F = Q^2/2\epsilon_0A = C^2V^2/2\epsilon_0A \).
2. Using \( \epsilon_0A = Cd \): \( F = CV^2/2d \).

Teacher's Note:
1. Replace ε₀A by C d.

Question 11: A capacitance of value 8 µF charged to 50 V is connected with another capacitance of value 4 µF charged to 100 V, in such a way that plates of similar charges are connected together. The total energy in multiples of 10⁻² J before joining and after joining will be
(a) 1.5 and 1.33
(b) 1.33 and 1.5
(c) 3.0 and 2.67
(d) 2.67 and 3.0
Show Answer & Explanation

Answer: (c) 3.0 and 2.67

Explanation:
1. Before: \( \tfrac12(8)(50)^2 + \tfrac12(4)(100)^2 = 10000 + 20000 = 30000 \) µJ = 3.0 × 10⁻² J.
2. Charge = 400 + 400 = 800 µC, \( C = 12 \) µF, \( V = 66.7 \) V.
3. After: \( Q^2/2C = (800)^2/24 = 26667 \) µJ = 2.67 × 10⁻² J.

Teacher's Note:
1. Total charge is conserved, not energy.
2. Energy always falls on sharing.

Question 12: A 3 µF capacitor is charged to a potential of 100 V and a 2 µF capacitor is charged to 100 V. The capacitors are then connected in parallel with plates of opposite polarities joined together. What amount of charge will flow when the plates are so connected?
(a) 1300 µC
(b) 800 µC
(c) 600 µC
(d) 240 µC
Show Answer & Explanation

Answer: (d) 240 µC

Explanation:
1. Charges: 300 µC and 200 µC, opposite. Net charge = 100 µC on 5 µF, so \( V = 20 \) V.
2. The 3 µF capacitor ends with 60 µC, so 240 µC of its charge moved.

Teacher's Note:
1. Net charge is the difference.
2. Charge that flows is the change on one capacitor.

Question 13: An air capacitor of capacity C = 10 µF is connected to a constant voltage battery of 12 V. Now the space between the plates is filled with a liquid of dielectric constant 3. The additional charge that flows now from the battery to the capacitor is
(a) 120 µC
(b) 600 µC
(c) 480 µC
(d) 240 µC
Show Answer & Explanation

Answer: (d) 240 µC

Explanation:
1. Initial charge = 10 × 12 = 120 µC.
2. With the liquid, \( C' = 30 \) µF and Q' = 360 µC.
3. Additional charge = 240 µC.

Teacher's Note:
1. The battery keeps V constant, so charge rises K times.

Question 14: In a capacitor of capacitance 20 µF the distance between the plates is 2 mm. If a dielectric slab of thickness 2 mm and dielectric constant 2 is inserted between the plates, then the new capacitance will be
(a) 22 µF
(b) 26.6 µF
(c) 52.2 µF
(d) 40 µF
Show Answer & Explanation

Answer: (d) 40 µF

Explanation:
1. The slab fills the whole gap, so capacitance is multiplied by K = 2.
2. New capacitance = 40 µF.

Teacher's Note:
1. A full-gap slab multiplies C by K.

Question 15: Two capacitors C₁ = 4 µF and C₂ = 4 µF are connected in series. This combination is then connected in parallel to a third capacitor C₃ = 4 µF. This arrangement is then connected to a battery of emf 4 V. How much energy is given by the battery in charging the capacitors?
(a) \( 96\times10^{-6} \) J
(b) \( 48\times10^{-6} \) J
(c) \( (32/3)\times10^{-6} \) J
(d) \( (16/3)\times10^{-6} \) J
Show Answer & Explanation

Answer: (a) \( 96\times10^{-6} \) J

Explanation:
1. C₁ and C₂ in series give 2 µF; in parallel with C₃ this gives 6 µF.
2. Energy given by the battery \( = CV^2 = 6\times10^{-6}\times16 = 96\times10^{-6} \) J.

Teacher's Note:
1. Battery energy is QV. Only half of it is stored.

Question 16: A slab of copper of thickness d/3 is introduced between the plates of a parallel plate capacitor where d is the separation between its two plates. If the capacity of the capacitor without the copper slab is C and with copper slab is C′ then C′/C is
(a) \( \sqrt2 \)
(b) 2
(c) 3/2
(d) \( 1/\sqrt2 \)
Show Answer & Explanation

Answer: (c) 3/2

Explanation:
1. \( C' = \epsilon_0A/(d - d/3) = 3\epsilon_0A/2d \).
2. \( C'/C = 3/2 \).

Teacher's Note:
1. A conductor slab effectively removes its thickness from the gap.

Question 17: Three parallel plates of the same metal and same area are placed between the plates of a parallel plate capacitor of capacity C. If the thickness of each plate is equal to 1/5th of the distance between the plates of the original capacitor then the capacity of the new capacitor is
(a) \( \tfrac53C \)
(b) \( \tfrac52C \)
(c) \( \tfrac3{10}C \)
(d) \( \tfrac{10}3C \)
Show Answer & Explanation

Answer: (b) \( \tfrac52C \)

Explanation:
1. Metal thickness removed from the gap = 3 × d/5 = 3d/5.
2. Effective gap = 2d/5, so \( C' = 5C/2 \).

Teacher's Note:
1. Each metal plate removes its own thickness from the effective gap.

Question 18: When a capacitor of value 200 µF charged to 200 V is discharged separately through resistances of 2 ohms and 8 ohms, then heat produced in joule respectively will be
(a) 4 and 16
(b) 16 and 4
(c) 4 and 8
(d) 4 and 4
Show Answer & Explanation

Answer: (d) 4 and 4

Explanation:
1. Total heat equals the initial energy \( \tfrac12CV^2 = \tfrac12(200\times10^{-6})(200)^2 = 4 \) J.
2. It does not depend on the resistance, so 4 J in both cases.

Teacher's Note:
1. All stored energy turns into heat.

Question 19: For a capacitor the graph of potential difference V (vertical axis) against charge q (horizontal axis) is a straight line through the origin. The area of the triangle bounded by the line, the q axis and the ordinate at a given q represents
(a) electric field between the plates
(b) electric flux between the plates
(c) energy density
(d) energy stored by the capacitor
Show Answer & Explanation

Answer: (d) energy stored by the capacitor

Explanation:
1. The area is \( \tfrac12qV \).
2. This is the energy stored in the capacitor.

Teacher's Note:
1. Work done in charging is the area under the V–q graph.

Question 20: A charged parallel plate capacitor of distance d has U₀ energy. A slab of dielectric constant K and thickness d is then introduced between the plates of the capacitor (battery disconnected). The new energy of the system is given by
(a) \( KU_0 \)
(b) \( K^2U_0 \)
(c) \( U_0/K \)
(d) \( U_0/K^2 \)
Show Answer & Explanation

Answer: (c) \( U_0/K \)

Explanation:
1. Q is constant and capacitance becomes KC.
2. \( U = Q^2/2KC = U_0/K \).

Teacher's Note:
1. With Q fixed, energy varies inversely with C.

Question 21: The energy and capacity of a charged parallel plate capacitor are U and C respectively. Now a dielectric slab of ε_r = 6 is inserted in it. Its energy and capacity become (assuming charge remains constant)
(a) 6U, 6C
(b) U, C
(c) U/6, 6C
(d) U, 6C
Show Answer & Explanation

Answer: (c) U/6, 6C

Explanation:
1. Capacity becomes 6C.
2. With Q constant, \( U = Q^2/2C \) becomes U/6.

Teacher's Note:
1. Capacity multiplies by K and energy divides by K.

Question 22: The distance between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of dielectric constant K and thickness t is placed between the plates, its capacity becomes
(a) \( \dfrac{\epsilon_0A}{d+t(1-1/K)} \)
(b) \( \dfrac{\epsilon_0A}{d+t(1+1/K)} \)
(c) \( \dfrac{\epsilon_0A}{d-t(1+1/K)} \)
(d) \( \dfrac{\epsilon_0A}{d-t(1-1/K)} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{\epsilon_0A}{d-t(1-1/K)} \)

Explanation:
1. Two layers in series: air of thickness d − t and slab of thickness t.
2. \( C = \epsilon_0A/(d - t + t/K) = \epsilon_0A/[d - t(1-1/K)] \).

Teacher's Note:
1. The slab counts as a gap of t/K.

Question 23: A parallel plate capacitor is connected with a battery whose potential difference remains constant. If the plates of the capacitor are shifted apart then the intensity of electric field
(a) decreases and charge on plates also decreases
(b) remains constant but charge on plates decreases
(c) remains constant but charge on the plates increases
(d) increases but charge on the plates decreases
Show Answer & Explanation

Answer: (a) decreases and charge on plates also decreases

Explanation:
1. V is fixed. \( E = V/d \) decreases as d increases.
2. C decreases and \( Q = CV \) decreases.

Teacher's Note:
1. The battery holds V constant.

Question 24: A parallel plate air capacitor has capacitance C. Its plates are vertical and it is dipped so that half of the area of the plates is in contact with a liquid of dielectric constant 5 (the liquid fills the whole gap over that half). The percentage increase in the capacitance will be
(a) 400%
(b) 66.6%
(c) 33.3%
(d) 200%
Show Answer & Explanation

Answer: (d) 200%

Explanation:
1. The two halves act in parallel: \( C' = C/2 + 5C/2 = 3C \).
2. Increase = 200%.

Teacher's Note:
1. Side-by-side regions add in parallel.

Question 25: Two conducting spheres of radii R₁ and R₂ are charged with charges Q₁ and Q₂ respectively. On bringing them in contact there is
(a) no change in the energy of the system
(b) an increase in the energy of the system if \( Q_1R_2\ne Q_2R_1 \)
(c) always a decrease in the energy of the system
(d) a decrease in the energy of the system if \( Q_1R_2\ne Q_2R_1 \)
Show Answer & Explanation

Answer: (d) a decrease in the energy of the system if \( Q_1R_2\ne Q_2R_1 \)

Explanation:
1. Charge flows until potentials are equal, i.e. \( Q_1/R_1 = Q_2/R_2 \).
2. If the potentials were already equal, nothing flows. Otherwise energy is lost as heat, so the energy decreases when \( Q_1R_2\ne Q_2R_1 \).

Teacher's Note:
1. Energy never increases on joining conductors.

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