The capacitance of a parallel plate capacitor can be increased by
(A) inserting a dielectric material between the plates.
(B) decreasing the distance between the plates.
(C) increasing the area of cross-section of the plates.
(D) connecting another identical parallel plate capacitor in series with it.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (C) only
The capacitance of a parallel plate capacitor filled with a medium of dielectric constant \( K \) is
\( \displaystyle C = \frac{K\varepsilon_{0}A}{d} \)
(A) inserting a dielectric material between the plates. — Correct. \( K \gt 1 \), so \( C \) increases \( K \) times.
(B) decreasing the distance between the plates. — Correct. \( C \propto 1/d \).
(C) increasing the area of cross-section of the plates. — Correct. \( C \propto A \).
(D) connecting another identical parallel plate capacitor in series with it. — Incorrect. Two equal capacitors in series give \( C/2 \); the capacitance would increase (to \( 2C \)) only in parallel.
NCERT (2026-27):
Effect of Dielectric on Capacitance – “… The capacitance C0 in this case is
\( \displaystyle C_0 = \frac{Q}{V_0} = \varepsilon_0 \frac{A}{d} \qquad (2.46) \)”
Effect of Dielectric on Capacitance – “… The capacitance C, with dielectric between the plates, is then
\( \displaystyle C = \frac{Q}{V} = \frac{\varepsilon_0 K A}{d} \qquad (2.51) \)”
Effect of Dielectric on Capacitance – “… Thus, the dielectric constant of a substance is the factor (>1) by which the capacitance increases from its vacuum value, when the dielectric is inserted fully between the plates of a capacitor.”