Question

As shown in fig., two identical capacitors are connected to a battery of $V\text{volts}$ in parallel. When capacitors are fully charged their energy is $U_1$. If the key $K$ is opened and a material of dielectric constant ${\epsilon }_r=2$ is inserted in each capacitors their stored energy is now $U_2.$ Then, $\frac{U_1}{U_2}$ will be

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

Answer 1

The correct option is C $\frac{4}{5}$

The given circuit is,


Initially, when the switch is closed both the capacitors A and B are in parallel and therefore, the energy stored in the capacitor is,

$U_1=2\times \frac{1}{2}C{V}^2...\left(1\right)$

When switch S is opened, B gets disconnected from the battery. The capacitor B is now isolated and the charge on an isolated capacitor remains constant. On the other hand A remains connected to the battery. Hence, potential $V$ remains constant on it.

When the capacitors are filled with a dielectric, their capacitance increase to $KC$ therefore, energy stored in B changes to $\frac{Q^2}{2KC}$

Where $Q=CV$ is the charge on B, which remains constant.

Thus the final total energy stored in the capacitors is,

$U_2=\frac{1}{2}\frac{(CV{)}^2}{KC}+\frac{1}{2}KC{V}^2=\frac{1}{2}C{V}^2(K+\frac{1}{K})....\left(2\right)$

Form equation $\left(1\right)$ and $\left(2\right)$ we find;

$\frac{U_1}{U_2}=\frac{2K}{K^2+1}$

As $K=2$

$\frac{U_1}{U_2}=\frac{2\times 2}{(2{)}^2+1}=\frac{4}{5}$