The electrostatic force between the plates of an isolated parallel plate capacitor having charge Q and area of each plate A is:
Answer & explanation
Correct answer: option 1
$\text{Electric Field due to one plate of capacitor at other plate is } E = \frac{\sigma}{2\epsilon_0} = \frac{Q}{2\epsilon_0 A}$
$ \text{Force } F = QE = \frac{Q^2}{2\epsilon_0 A}$