The electrostatic force between the plates of an isolated parallel plate capacitor having charge Q and area of each plate A is:

\(\frac{Q^2}{2Aε_0}\) \(Q^22Aε_0\) \(\frac{σ}{2ε_0}\) \(\frac{Q}{2Aε_0}\)

\(\frac{Q^2}{2Aε_0}\)

$\text{Electric Field due to one plate of capacitor at other plate is } E = \frac{\sigma}{2\epsilon_0} = \frac{Q}{2\epsilon_0 A}$ $ \text{Force } F = QE = \frac{Q^2}{2\epsilon_0 A}$