The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is…
The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $\mathrm{C}$ is its initial capacitance, its final capacitance is equal to:
$2 \mathrm{C}$
$C / 2$
$4 \mathrm{C}$
$\mathrm{C} / 4$
Solution
The capacitance of parallel plate capacitor,
$\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{d}$
(where, the symbols have their usual meanings.)
Given, $\quad d^{\prime}=2 \mathrm{~d} \& \mathrm{~A}^{\prime}=\frac{\mathrm{A}}{2}$
Then, $\mathrm{C}^{\prime}=\frac{\varepsilon_0\left(\frac{\mathrm{A}}{2}\right)}{2 d}=\frac{\varepsilon_0 \mathrm{~A}}{4 d}=\frac{\mathrm{C}}{4}$
.