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Question

In a parallel-plate capacitor with plate area $A$ and charge $Q$ the force on one plate because of the charge on the other is equal to.
$\frac{Q^{2}}{ε_{0} A^{2}}$
$\frac{Q^{2}}{2 ε_{0} A^{2}}$
$\frac{Q^{2}}{ε_{0} A}$
$\frac{Q^{2}}{2 ε_{0} A}$

A

\frac{Q^{2}}{ε_{0} A}

B

\frac{Q^{2}}{2 ε_{0} A}

C

\frac{Q^{2}}{ε_{0} A^{2}}

D

\frac{Q^{2}}{2 ε_{0} A^{2}}

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