Two capacitors of capacitance \(3\ \mu\mathrm{F}\) and \(5\ \mu\mathrm{F}\) are each given a charge of \(30\ \mu\mathrm{C}\). The potential difference across the \(3\ \mu\mathrm{F}\) capacitor is more than that across the \(5\ \mu\mathrm{F}\) capacitor by
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(4\ \mathrm{V}\)
For a capacitor the ratio of charge to potential difference is fixed, \( C=Q/V \). For the same charge, the capacitor with the smaller capacitance has the larger potential difference.
\( \begin{aligned} V_1 &= \frac{Q}{C_1} \\ &= \frac{30\ \mu\mathrm{C}}{3\ \mu\mathrm{F}} \\ &= 10\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} V_2 &= \frac{Q}{C_2} \\ &= \frac{30\ \mu\mathrm{C}}{5\ \mu\mathrm{F}} \\ &= 6\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} V_1-V_2 &= 10-6 \\ &= 4\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Capacitors and Capacitance – “… Consequently, V is also proportional to Q, and the ratio Q/V is a constant:”
\( C=\frac{Q}{V} \)\( \qquad (2.38) \)
Capacitors and Capacitance – “Equation (2.38) shows that for large C, V is small for a given Q. …”