Q. The energy required to charge a parallel plate condenser of plate separation $d$ and plate area of cross-section $A$ such that uniform electric field between the plates is $E$,is
A) $\frac{1}{2}\frac{\varepsilon_{0}E^{2}}{Ad}$
B) $\frac{\varepsilon_{0}E^{2}}{Ad}$
C) $\varepsilon_{0}E^{2}Ad$
D) $\frac{1}{2}\varepsilon_{0}E^{2}Ad$
Explanation
(c) Energy given by the cell
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = CV^{2}$
Here, $C$ = capacitance of capacitor = $\frac{A\varepsilon_{0}}{d}$
$\ \ \ \ \ \ \ \ \ \ \ V =$potential difference across the plates = $Ed$
Therefore, $\ \ \ \ \ \ \ E = \left( \frac{A\varepsilon_{0}}{d} \right)(Ed)^{2} = A\varepsilon_{0}E^{2}d$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.