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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
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Verified Answer
The correct answer is:
$\frac{C V^2}{2 d}$
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.
$F=E_1 \times Q=E_1 \times C V=\frac{V}{2 d} \times C V=\frac{C V^2}{2 d}$
$E=\frac{V}{d}$
Then, electric field due to only one plate.
$E_1=\frac{V}{2 d}$
$\therefore$ force of one plate on another.
$F=E_1 \times Q=E_1 \times C V=\frac{V}{2 d} \times C V=\frac{C V^2}{2 d}$
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