The energy required to charge a parallel plate condenser of plate separation d and plate area of…

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
  1. $\frac{1}{2} \varepsilon_0 E^2 / A d$
  2. $\varepsilon_0 E^2 / A d$
  3. $\varepsilon_0 E^2 A d$
  4. $\frac{1}{2} \varepsilon_0 E^2 A d$

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,
$\begin{aligned}
E & =\left(\frac{A \varepsilon_0}{d}\right)(E d)^2 \\
& =A \varepsilon_0 E^2 d
\end{aligned}$

Asked in: NEET 2008 (Screening)

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