If $Q$ is the charge on the plates of a capacitor of capacitance $C, V$ the potential difference between the plates, $A$ the area of each plate and $d $ the distance between the plates, the force of attraction between the plates is
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$F_{B}=Q\left(\frac{\sigma}{2 \in 0}\right)=\frac{Q^{2}}{2 \in 0 A}$
$F_{B}=\frac{Q^{2}}{2 \in 0 A} \frac{Q^{2}}{\frac{2 \epsilon_{0} A}{d} \times d}=\frac{Q^{2}}{2 C \times d}=\frac{1}{2} \frac{C V^{2}}{d}$
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