Question

Plates of a parallel plate capacitor $(C=6nF)$ carries charge $Q_1=12nC$ and $Q_2=6nC$, then energy stored in capacitor is

• A. $\frac{4}{5}nJ$
• B. $\frac{3}{4}nJ$
• C. $\frac{2}{3}nJ$
• D. $\frac{1}{3}nJ$

Answer 1

Answer: (B)

$\frac{3}{4}nJ$


At equilibrium,

$E$ inside the conductor with charge $Q_1$ is $=0$. So,

$\frac{Q_1-x}{2A{ϵ}_0}-\frac{x}{2A{ϵ}_0}+\frac{x}{2A{ϵ}_0}-\left(\frac{Q_2+x}{2A{ϵ}_0}\right)=0$

$Q_1-x-Q_2-x=0$

$x=\frac{Q_1-Q_2}{2}$

$\Delta V$ between plates $=\frac{x}{C}=\frac{Q_1-Q_2}{2C}$
$=\frac{12-6}{2\times 6}=0.5volt$
$V=\frac{1}{2}.C\Delta V^2=\frac{1}{2}\times 6\times (\frac{1}{2}{)}^2=\frac{6}{8}=\frac{3}{4}nJ$