Q116 / 150 · MHT CET 2008 · 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
$$
\begin{aligned}
& E=C V^{2} \\
\text { Here, } C &=\text { capacitance of condenser } \\
&=\frac{A \varepsilon_{0}}{d}
\end{aligned}
$$
$V=$ potential difference across the plates $=E d$
$$
\text { Therefore, } E=\left(\frac{A \varepsilon_{0}}{d}\right)(E d)^{2}=A \varepsilon_{0} E^{2} d
$$
Hence, the correct answer is option: 'C'.