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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 plates, 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:
$E_T=\frac{V}{d}$
Then, electric field due to only one plate:
$E_1=\frac{V}{2 d}$
$\therefore$ force of one plate on another:
$\mathrm{F}=\mathrm{E}_1 \times \mathrm{QF}=\mathrm{E}_1 \times \mathrm{CV}=\frac{\mathrm{V}}{2 \mathrm{~d}} \times \mathrm{CV}=\frac{\mathrm{CV}^2}{2 \mathrm{~d}}$
$E_T=\frac{V}{d}$
Then, electric field due to only one plate:
$E_1=\frac{V}{2 d}$
$\therefore$ force of one plate on another:
$\mathrm{F}=\mathrm{E}_1 \times \mathrm{QF}=\mathrm{E}_1 \times \mathrm{CV}=\frac{\mathrm{V}}{2 \mathrm{~d}} \times \mathrm{CV}=\frac{\mathrm{CV}^2}{2 \mathrm{~d}}$
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