Consider a parallel-plate capacitor with plate area A and plate separation d. Suppose a positive charge +q is given to one plate and a negative charge −q to the other plate. The magnitude of the force between the plates of the capacitor will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 0.5\,q^{2}A^{-1}\varepsilon_{0}^{-1} \)
The force on one plate is due to the field of the other plate only, not the total field between the plates.
The field of one plate with surface charge density \( \sigma = q/A \) is
\( \displaystyle E_1 = \frac{\sigma}{2\varepsilon_0} = \frac{q}{2A\varepsilon_0} \)
Force on the other plate (charge q):
\( \displaystyle F = qE_1 = \frac{q^{2}}{2A\varepsilon_0} = 0.5\,q^{2}A^{-1}\varepsilon_0^{-1} \)
The force is independent of d because the field of a large plane sheet does not depend on distance.
Note: This question is from beyond the NCERT textbook and it appears to be out of the CUET 2026 Physics syllabus.