The potential difference across a capacitor is raised from 20 V to 50 V. As a result, the charge on the capacitor increases by \(90\ \mu\mathrm{C}\). The charge on the capacitor at 50 V is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(150\ \mu\mathrm{C}\)
The capacitance does not depend on the charge or the voltage. So the charge grows in step with the voltage, and the change in charge is \( \Delta Q=C\,\Delta V \). This gives \( C \) first.
\( \begin{aligned} C &= \frac{\Delta Q}{\Delta V} \\ &= \frac{90\ \mu\mathrm{C}}{(50-20)\ \mathrm{V}} \\ &= 3\ \mu\mathrm{F} \end{aligned} \)
\( \begin{aligned} Q &= CV \\ &= 3\ \mu\mathrm{F}\times50\ \mathrm{V} \\ &= 150\ \mu\mathrm{C} \end{aligned} \)
Therefore, the charge at 50 V is \( 150\ \mu\mathrm{C} \). At 20 V it was \( 60\ \mu\mathrm{C} \), which is \( 90\ \mu\mathrm{C} \) less.
NCERT (2026-27):
Capacitors and Capacitance – “… The constant C is called the capacitance of the capacitor. C is independent of Q or V, as stated above. …”
Summary – “… Its capacitance is defined by C = Q/V, where Q and –Q are the charges on the two conductors and V is the potential difference between them. …”