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
The correct answer is Option (1) → \(\frac{Q^2}{2 A \varepsilon_0} \)
Electric field due to plate, $E = \frac{Q}{2\epsilon_0 A}$
$⇒F = QE = \frac{Q^2}{2\epsilon_0 A}$