CUET Preparation Today
The electrostatic force between the plates of an isolated parallel plate capacitor having charge Q and area of each plate A is: |
\(\frac{Q^2}{2 A \varepsilon_0} \) \(Q^2 2 A \varepsilon_0 \) \(\frac{\sigma}{2 \varepsilon_0} \) \(\frac{Q}{2 A \varepsilon_0} \) |
\(\frac{Q^2}{2 A \varepsilon_0} \) |
The correct answer is Option (1) → \(\frac{Q^2}{2 A \varepsilon_0} \) Electric field due to plate, $E = \frac{Q}{2\epsilon_0 A}$ $⇒F = QE = \frac{Q^2}{2\epsilon_0 A}$ |
