A parallel plate air capacitor has a capacity '\( C \)', distance of separation between its plates is '\( x \)' and potential difference applied across the plates is \( V \). What is the force of attraction between the plates of the parallel plate air capacitor?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{CV^{2}}{2x} \)
Each plate is pulled by the field of the other plate alone, not by the total field between the plates. A plate with surface charge density \( \sigma \) produces a field \( \sigma/2\varepsilon_0 \), half the field \( \sigma/\varepsilon_0 \) between the plates. So the force on a plate of charge \( Q \) and area \( A \) is
\( \displaystyle F = Q \times \frac{\sigma}{2\varepsilon_0} = \frac{Q^{2}}{2\varepsilon_0 A} \)
Writing it in terms of \( C \), \( V \) and \( x \), with \( Q = CV \) and \( C = \varepsilon_0 A/x \), that is \( \varepsilon_0 A = Cx \):
\( \displaystyle F = \frac{C^{2}V^{2}}{2\,Cx} = \frac{CV^{2}}{2x} \)
Shortcut: Energy method: the stored energy at constant charge is \( U = Q^{2}x/(2\varepsilon_0 A) \), and the attraction is its rate of change with separation:
\( \displaystyle F = \frac{dU}{dx} = \frac{Q^{2}}{2\varepsilon_0 A} = \frac{CV^{2}}{2x} \)
Note: A similar question was also asked in CUET (UG) 2023 on 08-06-2023 (Slot 3).