JEE Physics Electrostatic Potential and Capacitance MCQs Set 04

Physics Objective Questions and Answers: Electrostatic Potential and Capacitance

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Multiple Choice Questions

Question 1: The capacitance C of a capacitor is
(a) proportional to the charge on it
(b) independent of the charge and the potential difference
(c) inversely proportional to the potential difference only
(d) independent of the geometry of the conductors
Show Answer & Explanation

Answer: (b) independent of the charge and the potential difference

Explanation:
1. \( C = Q/V \) is a constant fixed by geometry and the medium: if Q is doubled, V also doubles.

Teacher's Note:
1. Capacitance depends on shape, size, separation and the medium.

Question 2: To increase the charge on a plate of a capacitor means
(a) to increase the potential difference between the plates
(b) to decrease the potential difference
(c) to decrease the capacitance
(d) to increase the capacitance
Show Answer & Explanation

Answer: (a) to increase the potential difference between the plates

Explanation:
1. \( Q = CV \) and C is fixed, so more charge needs more potential difference.

Teacher's Note:
1. C does not change with charge.

Question 3: The net charge on a charged capacitor is
(a) 2q
(b) q/2
(c) infinite
(d) zero
Show Answer & Explanation

Answer: (d) zero

Explanation:
1. The plates carry +q and -q, so the total charge is zero.

Teacher's Note:
1. The capacitor stores energy, not net charge.

Question 4: If the Earth has volume V and surface area A, its capacitance is
(a) \( 4\pi\varepsilon_0\dfrac{A}{V} \)
(b) \( 4\pi\varepsilon_0\dfrac{V}{A} \)
(c) \( 12\pi\varepsilon_0\dfrac{V}{A} \)
(d) \( 12\pi\varepsilon_0\dfrac{A}{V} \)
Show Answer & Explanation

Answer: (c) \( 12\pi\varepsilon_0\dfrac{V}{A} \)

Explanation:
1. \( V = \frac43\pi R^3 \) and \( A = 4\pi R^2 \), so \( R = 3V/A \).
2. \( C = 4\pi\varepsilon_0R = 12\pi\varepsilon_0V/A \).

Teacher's Note:
1. Use the volume to area ratio R/3.

Question 5: A sphere has maximum circumference 2 m. Its capacitance when placed in water (dielectric constant 81) is nearly
(a) 27.65 pF
(b) 2865 pF
(c) 2385 pF
(d) 236.5 pF
Show Answer & Explanation

Answer: (b) 2865 pF

Explanation:
1. \( 2\pi R = 2 \Rightarrow R = 0.318 \) m.
2. \( C = 4\pi\varepsilon_0KR = \dfrac{81\times0.318}{9\times10^9} = 2.86\times10^{-9} \) F \( = 2865 \) pF.

Teacher's Note:
1. In air the value would be 35.4 pF.

Question 6: Two parallel plates are connected to a 300 V battery and each plate collects a charge of 1 microcoulomb. The energy supplied by the battery is
(a) \( 3\times10^{-4} \) J
(b) \( 6\times10^{-4} \) J
(c) \( 1.5\times10^{-4} \) J
(d) \( 4.5\times10^{-4} \) J
Show Answer & Explanation

Answer: (a) \( 3\times10^{-4} \) J

Explanation:
1. Work done by the battery \( = QV = 10^{-6}\times300 = 3\times10^{-4} \) J.
2. Half of it, \( 1.5\times10^{-4} \) J, is stored in the capacitor.

Teacher's Note:
1. The battery supplies QV, the capacitor stores QV/2.

Question 7: A 200 microfarad capacitor charged to 200 V is discharged separately through resistances of 2 ohm and 8 ohm. The heat produced is respectively
(a) 4 J and 16 J
(b) 16 J and 4 J
(c) 4 J and 8 J
(d) 4 J and 4 J
Show Answer & Explanation

Answer: (d) 4 J and 4 J

Explanation:
1. Energy stored \( = \frac12CV^2 = \frac12\times200\times10^{-6}\times200^2 = 4 \) J.
2. All of it is dissipated whatever the resistance.

Teacher's Note:
1. Heat depends on the stored energy, not on R.

Question 8: The potential to which a conductor is raised depends on
(a) only the amount of charge
(b) only the geometry
(c) both the amount of charge and the geometry and size of the conductor
(d) neither
Show Answer & Explanation

Answer: (c) both the amount of charge and the geometry and size of the conductor

Explanation:
1. \( V = Q/C \) and C depends on geometry and size.

Teacher's Note:
1. This is the basis of capacitance.

Question 9: On charging an uncharged capacitor with a battery,
(a) all the energy supplied is stored
(b) half the energy supplied by the battery is stored in the capacitor
(c) energy stored depends on the charging time
(d) energy stored depends on the resistance in the circuit
Show Answer & Explanation

Answer: (b) half the energy supplied by the battery is stored in the capacitor

Explanation:
1. Battery supplies QV; stored energy is QV/2; the rest is heat in the connecting wires.

Teacher's Note:
1. This is independent of the resistance.

Question 10: The capacity of a parallel plate condenser depends on
(a) the separation between the plates
(b) the type of metal used
(c) the thickness of the plates
(d) the potential difference applied across the plates
Show Answer & Explanation

Answer: (a) the separation between the plates

Explanation:
1. \( C = \varepsilon_0A/d \) depends on area, separation and the medium.

Teacher's Note:
1. It does not depend on the material of the plates.

Question 11: A parallel plate capacitor has plates of area 400 cm\(^2\) separated by 2 mm of air. The charge on the plates for a potential difference of 200 V is nearly
(a) \( 3.54\times10^{-6} \) C
(b) \( 3.54\times10^{-10} \) C
(c) \( 1.77\times10^{-8} \) C
(d) \( 3.54\times10^{-8} \) C
Show Answer & Explanation

Answer: (d) \( 3.54\times10^{-8} \) C

Explanation:
1. \( C = \dfrac{\varepsilon_0A}{d} = \dfrac{8.85\times10^{-12}\times0.04}{2\times10^{-3}} = 1.77\times10^{-10} \) F.
2. \( Q = CV = 1.77\times10^{-10}\times200 = 3.54\times10^{-8} \) C.

Teacher's Note:
1. Convert cm squared to m squared (400 cm2 = 0.04 m2).

Question 12: The distance between the plates of a circular parallel plate capacitor of diameter 40 mm, whose capacity equals that of a metallic sphere of radius 1 m, is
(a) 0.01 mm
(b) 1.0 mm
(c) 0.1 mm
(d) 10 mm
Show Answer & Explanation

Answer: (c) 0.1 mm

Explanation:
1. \( \dfrac{\varepsilon_0\pi r^2}{d} = 4\pi\varepsilon_0R \Rightarrow d = \dfrac{r^2}{4R} \).
2. \( d = \dfrac{(0.02)^2}{4\times1} = 10^{-4} \) m \( = 0.1 \) mm.

Teacher's Note:
1. Radius of plate is half the diameter.

Question 13: The energy density in a parallel plate capacitor is \( 2.1\times10^{-9} \) J m\(^{-3}\). The electric field between the plates is nearly
(a) 2.1 N C\(^{-1}\)
(b) 21.6 N C\(^{-1}\)
(c) 72 N C\(^{-1}\)
(d) 8.4 N C\(^{-1}\)
Show Answer & Explanation

Answer: (b) 21.6 N C\(^{-1}\)

Explanation:
1. \( u = \frac12\varepsilon_0E^2 \Rightarrow E = \sqrt{\dfrac{2u}{\varepsilon_0}} = \sqrt{\dfrac{4.2\times10^{-9}}{8.85\times10^{-12}}} = 21.8 \) N C\(^{-1}\).

Teacher's Note:
1. The closest option is 21.6.

Question 14: A charged parallel plate capacitor (battery removed) has energy U0. A dielectric slab of constant K fully filling the gap is introduced. The new energy is
(a) \( U_0/K \)
(b) \( KU_0 \)
(c) \( K^2U_0 \)
(d) \( U_0/K^2 \)
Show Answer & Explanation

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

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

Teacher's Note:
1. For an isolated capacitor the energy falls when a dielectric is introduced.

Question 15: The energy and capacity of a charged parallel plate capacitor are U and C. A dielectric slab of dielectric constant 6 is inserted with the charge on the plates remaining constant. Energy and capacity become
(a) 6U and 6C
(b) U and C
(c) U and 6C
(d) U/6 and 6C
Show Answer & Explanation

Answer: (d) U/6 and 6C

Explanation:
1. Capacity becomes KC = 6C.
2. Energy \( = Q^2/2(6C) = U/6 \).

Teacher's Note:
1. Charge constant: energy falls by K.

Question 16: The distance between the plates of a parallel plate capacitor is d and the area is A. A slab of dielectric constant K and thickness t (less than d) is placed between the plates. The capacitance becomes
(a) \( \dfrac{\varepsilon_0A}{d+t\left(1-\frac1K\right)} \)
(b) \( \dfrac{\varepsilon_0A}{d+t\left(1+\frac1K\right)} \)
(c) \( \dfrac{\varepsilon_0A}{d-t\left(1-\frac1K\right)} \)
(d) \( \dfrac{\varepsilon_0A}{d-t\left(1+\frac1K\right)} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{\varepsilon_0A}{d-t\left(1-\frac1K\right)} \)

Explanation:
1. Air gap \( d-t \) in series with the dielectric layer of effective gap \( t/K \).
2. \( C = \dfrac{\varepsilon_0A}{d-t+t/K} \).

Teacher's Note:
1. The slab reduces the effective separation.

Question 17: A dielectric slab is placed between the plates of a parallel plate capacitor which is connected to a battery. Compared with the earlier charge, the charge on the plates
(a) is less
(b) is more
(c) is the same
(d) depends on the nature of the material
Show Answer & Explanation

Answer: (b) is more

Explanation:
1. V is constant and C increases, so \( Q = CV \) increases.

Teacher's Note:
1. The battery supplies more charge.

Question 18: A glass slab is put within the plates of a charged, isolated parallel plate capacitor. Which quantity does NOT change?
(a) charge
(b) energy of the capacitor
(c) capacity
(d) intensity of the electric field
Show Answer & Explanation

Answer: (a) charge

Explanation:
1. The capacitor is isolated, so charge cannot leave. The capacity increases, the voltage and field fall and the energy falls.

Teacher's Note:
1. Isolated means constant charge.

Question 19: A parallel plate capacitor is connected to a battery and a dielectric slab is inserted between the plates. Which quantity increases?
(a) potential difference
(b) electric field
(c) emf of the battery
(d) stored energy
Show Answer & Explanation

Answer: (d) stored energy

Explanation:
1. V is fixed by the battery, so E is constant. C and Q increase, so \( U = \frac12CV^2 \) increases.

Teacher's Note:
1. Battery connected means constant potential.

Question 20: A parallel plate capacitor is charged by a battery, the battery is disconnected and a dielectric plate is inserted. Which of the following is NOT true?
(a) The potential difference decreases
(b) The electric field decreases
(c) The stored energy increases
(d) The capacitance increases
Show Answer & Explanation

Answer: (c) The stored energy increases

Explanation:
1. Q is constant, C increases by K, so V = Q/C and E decrease, and \( U = Q^2/2C \) decreases.

Teacher's Note:
1. Stored energy decreases with a dielectric at constant charge.

Question 21: A parallel plate capacitor is charged and the battery is disconnected. The distance between the plates is then decreased. Which is correct?
(a) The electric field changes
(b) The stored energy decreases
(c) The potential difference increases
(d) The capacitance decreases
Show Answer & Explanation

Answer: (b) The stored energy decreases

Explanation:
1. \( E = \sigma/\varepsilon_0 \) is unchanged. \( C = \varepsilon_0A/d \) increases, \( V = Qd/\varepsilon_0A \) decreases, and \( U = Q^2/2C \) decreases.

Teacher's Note:
1. Constant charge: field constant, energy falls.

Question 22: A parallel plate capacitor is connected to a battery of constant potential difference. If the plates are pulled apart, the electric field between them
(a) decreases and the charge on the plates also decreases
(b) remains constant and charge also remains constant
(c) remains constant but charge decreases
(d) increases and charge decreases
Show Answer & Explanation

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

Explanation:
1. \( E = V/d \) decreases as d increases and \( Q = CV \) decreases because C falls.

Teacher's Note:
1. Battery connected: V is constant.

Question 23: A parallel plate capacitor is charged and the battery removed. The distance between the plates is increased using insulating handles. Then
(a) the capacitance increases
(b) the charge on the capacitor increases
(c) the stored energy decreases
(d) the potential difference between the plates increases
Show Answer & Explanation

Answer: (d) the potential difference between the plates increases

Explanation:
1. Q is constant. \( C = \varepsilon_0A/d \) falls, so \( V = Q/C \) rises and \( U = Q^2/2C \) rises.

Teacher's Note:
1. Work done pulling the plates apart goes into the energy.

Question 24: 1 \( \mu F \) capacitors, each of which breaks down above 500 V, are available. The least number of capacitors needed to make a 3 \( \mu F \) capacitor able to withstand 2000 V is
(a) 4
(b) 12
(c) 48
(d) 16
Show Answer & Explanation

Answer: (c) 48

Explanation:
1. Each row must have 4 capacitors in series to withstand 2000 V; one row gives \( 1/4 \) \( \mu F \).
2. For 3 \( \mu F \) we need 12 rows in parallel, so \( 4\times12 = 48 \) capacitors.

Teacher's Note:
1. Series for voltage, parallel for capacitance.

Question 25: A series combination of 0.1 \( \mu F \) and 1 \( \mu F \) capacitors is connected to a 500 V source. The potential difference across the 0.1 \( \mu F \) capacitor is nearly
(a) 50 V
(b) 454.5 V
(c) 500 V
(d) 45.5 V
Show Answer & Explanation

Answer: (b) 454.5 V

Explanation:
1. Charge is the same on both, so voltage is inversely proportional to capacitance.
2. \( V_{0.1} = 500\times\dfrac{1}{0.1+1} = 454.5 \) V.

Teacher's Note:
1. The smaller capacitor takes the larger voltage.

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