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