Question

A parallel plate capacitor is connected to a battery and a dielectric plate is inserted into the capacitor. Which quantity will increase?

• A. Potential difference across the capacitor
• B. Electric field
• C. Stored energy
• D. EMF of battery

Answer 1

The correct option is C Stored energy

The potential difference across the capacitor will be equal to the emf of the battery and will not change on inserting the dielectric.
Hence, option $\left(a\right)$ is incorrect.

Electric field is given by
$E=\frac{V}{d}$
As $V$ and $d$ are constant, $E$ will not change.
Hence, option $\left(b\right)$ is incorrect.

As the dielectric is inserted with the battery connected, $V$ is constant.
Initial energy stored in capacitor is $\frac{1}{2}C{V}^2$. And, final energy stored in capacitor (after inserting dielectric) will be $\frac{1}{2}\left(KC\right)V^2$

As $K>1$, energy stored in the capacitor will increase.
Hence, option $\left(c\right)$ is correct.

The emf of the battery does not depend on what circuit elements are connected in a circuit. So, it will not change with the insertion of the dielectric plate into the capacitor.
Hence, option $\left(d\right)$ is incorrect.