The force between the plates of a parallel plate capacitor of capacitance C, distance between plates d and potential difference V is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{CV^{2}}{2d} \)
The force on one plate is the charge on it multiplied by the electric field produced by the other plate alone. Each plate of charge density \( \sigma = Q/A \) produces a field \( \sigma/2\varepsilon_0 \), while the total field between the plates, \( E = \sigma/\varepsilon_0 = V/d \), is the sum of the two.
\( \displaystyle F = Q\,\frac{E}{2} = \frac{1}{2}\,Q\,\frac{V}{d} \)
Using \( Q = CV \):
\( \displaystyle F = \frac{1}{2}\,(CV)\,\frac{V}{d} = \frac{CV^{2}}{2d} \)
Note: A similar question was also asked in CUET (UG) 2023 on 23-05-2023 (Slot 3).