A parallel plate capacitor of capacitance \( 8.0\times10^{-11}\,\mathrm{F} \) is being charged by an external source. The charging current is constant and equal to 0.15 A. The rate of change of potential difference between the plates is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.875\times10^{9}\,\mathrm{V\,s^{-1}} \)
The charge on a capacitor is \( q=CV \), and the charging current is the rate at which this charge grows. With \( C \) fixed, the current fixes the rate of change of potential difference:
\( \begin{aligned} i &= \frac{dq}{dt} \\ &= C\,\frac{dV}{dt} \end{aligned} \)
Hence:
\( \begin{aligned} \frac{dV}{dt} &= \frac{i}{C} \\ &= \frac{0.15\,\mathrm{A}}{8.0\times10^{-11}\,\mathrm{F}} \\ &= 1.875\times10^{9}\,\mathrm{V\,s^{-1}} \end{aligned} \)