Q122 / 200 · NEET 2022 (Phase 2) · Physics

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:

A
$2 \mathrm{C}$
B
$C / 2$
C
$4 \mathrm{C}$
Correct answer
D
$\mathrm{C} / 4$

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, $\mathrm{C}^{\prime}=\frac{\varepsilon_0\left(\frac{\mathrm{A}}{2}\right)}{2 d}=\frac{\varepsilon_0 \mathrm{~A}}{4 d}=\frac{\mathrm{C}}{4}$

Hence, the correct answer is option: 'D'.

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