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) → \( \displaystyle\frac{Q^{2}}{2A\varepsilon_0} \)
Each plate is pulled by the field of the other plate only; a plate exerts no net force on itself. The field due to one plate, of surface charge density \( \sigma = Q/A \), is
\( \displaystyle E_{1} = \frac{\sigma}{2\varepsilon_0} = \frac{Q}{2A\varepsilon_0} \)
So the force on the other plate, which carries charge \( Q \), is
\( \displaystyle F = Q\,E_{1} = Q \times \frac{Q}{2A\varepsilon_0} = \frac{Q^{2}}{2A\varepsilon_0} \)