Question

An uncharged parallel plate capacitor of capacitance $C$ completely filled with a dielectric slab of dielectric constant $K$ is connected across a battery of emf $E$. Then heat generated in the wires till the capacitor is fully charged is

• A. $\frac{1}{2}KC{E}^2$
• B. $\frac{1}{2}(K+1)C{E}^2$
• C. $\frac{3}{2}(K-1)C{E}^2$
• D. $\frac{1}{2}(K-1)C{E}^2$

Answer 1

The correct option is A $\frac{1}{2}KC{E}^2$

The capacitance of the capacitor with the dielectric material will be $KC$.

Hence, final energy stored in capacitor will be

$U=\frac{1}{2}KC{E}^2$

The heat energy dissipated will be equal to the energy stored i.e. half of the work done by the battery.

$H=\frac{1}{2}KC{E}^2$

Hence, option $\left(a\right)$ is correct.

Key Concept - Heat dissipated = work done by battery - change in potential energy stored.