Q115 / 150 · MHT CET 2022 (10 Aug Shift 2) · Physics

The force between the plates of a parallel plate capacitor of capacitance $C$ and distance of separation of the plates $d$ with a potential difference $V$ between the plate is

Correct answer
A
$\frac{C V^2}{2 d}$
B
$\frac{C^2 V^2}{2 d^2}$
C
$\frac{C^2 V^2}{d^2}$
D
$\frac{V^2 d}{C}$

Solution

Total electric field between the plates of the capacitor. $E=\frac{V}{d}$ Then, electric field due to only one plate. $E_1=\frac{V}{2 d}$ $\therefore$ force of one plate on another. $F=E_1 \times Q=E_1 \times C V=\frac{V}{2 d} \times C V=\frac{C V^2}{2 d}$

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

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