Q. Force of attraction between the plates of a parallel plate capacitor is
A) $\frac{q^{2}}{2\varepsilon_{0}AK}$
B) $\frac{q^{2}}{\varepsilon_{0}AK}$
C) $\frac{q}{2\varepsilon_{0}A}$
D) $\frac{q^{2}}{2\varepsilon_{0}A^{2}K}$
Explanation
(a) Force on one plate due to another is
$F = qE = q \times \frac{\sigma}{2\varepsilon_{0}K} = q\left( \frac{q}{2AK\varepsilon_{0}} \right) = \frac{q^{2}}{2AK\varepsilon_{0}}$
[where $\frac{\sigma}{2\varepsilon_{0}K}$ is the electric field produced by one plate at the location of other]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.