Q109 / 200 · NEET 2008 (Screening) · Physics
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$\frac{1}{2} \varepsilon_0 E^2 / A d$
- B$\varepsilon_0 E^2 / A d$
- Correct answerC$\varepsilon_0 E^2 A d$
- D$\frac{1}{2} \varepsilon_0 E^2 A d$
Solution
Energy given by the cell
$E=C V^2$
Here, $C=$ capacitance of condenser $=\frac{A \varepsilon_0}{d}$ $V=$ potential difference across the plates $=E d$ Therefore,
$\begin{aligned}
E & =\left(\frac{A \varepsilon_0}{d}\right)(E d)^2 \\
& =A \varepsilon_0 E^2 d
\end{aligned}$
Hence, the correct answer is option: 'C'.