Question

A parallel plate capacitor has a uniform electric field $\overrightarrow{E}$ in the space between the plates. If the distance between the plates is $d$ and the area of each plate
is $A,$ the energy stored in the capacitor is
(${\epsilon }_0=$ permittivity of free space)

• A. $\frac{E^{2}Ad}{{\epsilon }_0}$
• B. $\frac{1}{2}{\epsilon }_0{E}^2$
• C. ${\epsilon }_{0}EAd$
• D. $\frac{1}{2}{\epsilon }_0{E}^{2}Ad$

Answer 1

The correct option is D $\frac{1}{2}{\epsilon }_0{E}^{2}Ad$


Energy density associated with an electric field is given by

$u=\frac{dU}{dV}=\frac{1}{2}{\epsilon }_0{E}^2$

$\Rightarrow dU=\frac{1}{2}{\epsilon }_0{E}^{2}dV$

Total energy stored in the space between the capacitor will be

$U=\int dU=\int \frac{1}{2}{\epsilon }_0{E}^{2}dV$

$U=\frac{1}{2}{\epsilon }_0{E}^2\int dV\text{[E is constant]}$

$\therefore U=\frac{1}{2}{\epsilon }_0{E}^{2}V=\frac{1}{2}{\epsilon }_0{E}^{2}Ad[\because V=Ad]$

Hence, option $\left(\text{D}\right)$ is correct.