Q. A parallel plate air capacitor has capacity 'C' distance of separation between plates is 'd' and potential difference 'V' is applied between the plates force of attraction between the plates of the parallel plate air capacitor is :
A) $\frac{C^2V^2}{2d^2}$
B) $\frac{C^2V^2}{2d}$
C) $\frac{CV^2}{2d}$
D) $\frac{CV^2}{d}$
Explanation
[Solution] $F = \frac{Q^2}{2\varepsilon_0 A}$ with Q = CV and $\varepsilon_0 A = Cd$ gives $F = \frac{CV^2}{2d}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.