Force of attraction between the plates of a parallel plate …

Physics Electrostatics Potential & Capacitance Capacitance Single Correct MCQ

Published by: CGP EDU Academic Team

Published on: September 12, 2026

Force of attraction between the plates of a parallel plate capacitor is

$\frac{q^{2}}{2 \varepsilon_{0} A K}$

$\frac{q^{2}}{\varepsilon_{0} A K}$

$\frac{q}{2 \epsilon_0 A}$

$\frac{q^{2}}{2 \varepsilon_{0} A^{2} K}$

Text Solution

Verified by Experts

The correct answer is:

Force on one plate due to another is

F = qE = $q \times \frac{\overline{\sigma}}{2 \epsilon_0 K}$ $=q\left(\frac{q}{2AK\varepsilon_0}\right)=\frac{q^2}{2AK\varepsilon_0}$

(where $\frac{\bar{\sigma}}{2 \epsilon_0 k}$ is the electric field produced by one plate at the location of other).

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