Question

The energy required to charge a parallel plate condenser (capacitor) of plate separation $d$ and plate area of cross-section $A$ such that the uniform electric field between the plates is $E$, is :

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

Answer 1

Answer: (C)

${\epsilon }_0{E}^{2}Ad$

Energy given by the cell
$E=C{V}^2$

Here, $C=$ capacitance of condenser
$=\frac{A{\epsilon }_0}{d}$

$V=$ potential difference across the plates $=Ed$

Therefore, $E=\left(\frac{A{\epsilon }_0}{d}\right){\left(Ed\right)}^2=A{\epsilon }_0{E}^{2}d$