Capacitors and Capacitance — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Capacitors and Capacitance MCQs with step-by-step solutions (6 questions). Part of Electrostatic Potential and Capacitance. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Capacitors and Capacitance · medium · numerical
A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now, an insulating slab of same area but thickness d/2 is inserted between the plates having dielectric constant K = 4. The ratio of new capacitance to its original capacitance will be
A. 2 : 1
B. 8 : 5 ✓ Correct
C. 6 : 5
D. 4 : 1
Solution: With a slab of thickness $t = \frac{d}{2}$, $K = 4$: $C = \frac{\varepsilon_0 A}{(d - t) + \frac{t}{K}} = \frac{\varepsilon_0 A}{\frac{d}{2} + \frac{d}{8}} = \frac{8}{5}\frac{\varepsilon_0 A}{d}$
$\frac{C}{C_0} = \frac{8}{5}$
Q2 — Capacitors and Capacitance · medium · numerical
The capacitance of a parallel plate capacitor with air as medium is 6 µF. With the introduction of a dielectric medium, the capacitance becomes 30 µF. The permittivity of the medium is ($\varepsilon_0 = 8.85 \times 10^{-12}$ C²N⁻¹m⁻²)
A. $1.77 \times 10^{-12}$ C²N⁻¹m⁻²
B. $0.44 \times 10^{-10}$ C²N⁻¹m⁻² ✓ Correct
C. 5.00 C²N⁻¹m⁻²
D. $0.44 \times 10^{-13}$ C²N⁻¹m⁻²
Solution: Dielectric constant $K = \frac{C_m}{C_0} = \frac{30}{6} = 5$.
$\varepsilon_m = K\varepsilon_0 = 5 \times 8.85 \times 10^{-12} = 0.44 \times 10^{-10}$ C²N⁻¹m⁻²
Q3 — Capacitors and Capacitance · medium · theory
The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is
A. proportional to the square root of the distance between the plates
B. linearly proportional to the distance between the plates
C. independent of the distance between the plates ✓ Correct
D. inversely proportional to the distance between the plates
Solution: $F = \frac{Q^2}{2Cd} = \frac{Q^2 d}{2\varepsilon_0 A d} = \frac{Q^2}{2\varepsilon_0 A}$ — the distance d cancels, so the force is independent of the plate separation.
Q4 — Capacitors and Capacitance · medium · numerical
A parallel plate air capacitor has capacity C, distance of separation between plates is d and potential difference V is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is
A. $\frac{C^2V^2}{2d}$
B. $\frac{CV^2}{2d}$ ✓ Correct
C. $\frac{CV^2}{d}$
D. $\frac{C^2V^2}{2d^2}$
Solution: $F = \frac{q^2}{2\varepsilon_0 A}$ with $q = CV$ and $C = \frac{\varepsilon_0 A}{d}$:
$F = \frac{C^2V^2}{2\varepsilon_0 A} = \frac{CV^2}{2d}$
Q5 — Capacitors and Capacitance · medium · theory
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 increased using an insulating handle. As a result the potential difference between the plates
A. decreases
B. does not change
C. becomes zero
D. increases ✓ Correct
Solution: With the battery disconnected the charge q is fixed. Increasing d decreases $C = \frac{\varepsilon_0 A}{d}$, and since $V = \frac{q}{C}$, the potential difference increases.
Q6 — Capacitors and Capacitance · easy · numerical
A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. If oil is removed, the capacitance of capacitor becomes
A. $\sqrt{2}C$
B. $2C$
C. $\frac{C}{\sqrt{2}}$
D. $\frac{C}{2}$ ✓ Correct
Solution: With oil $C = K C_0 = 2C_0$. Removing it gives $C_0 = \frac{C}{K} = \frac{C}{2}$