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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(a) 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$:$

$F=E_{1} \times Q F=E_{1} \times C V=\frac{V}{2 d} \times C V=\frac{C V^{2}}{2 d}$

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