Q. Assertion: A parallel plate capacitor is connected across battery through a key. A dielectric slab of dielectric constant $K$ is introduced between the plates. The energy which is stored becomes $K$ times.
Reason: The surface density of charge on the plate remains constant or unchanged.

Solution:

If a dielectric slab of dielectric constant $K$ is filled in between the plates of a condenser while charging it, the potential difference between the plates does not change but the capacity becomes $K$ times, therefore $V^{\prime}=V, C^{\prime}=K C$
$\therefore $ Energy stored in the capacitor
$U' =\frac{1}{2} C' V' ^{2} $
$=\frac{1}{2}(K C)\left(V^{2}\right) $
$=\left[\frac{1}{2} C V^{2}\right] K $
$=K U$
Thus, energy stored becomes $K$ times. Surface charge density
$\sigma'=\frac{q'}{A}=\frac{C' V'}{A} =\frac{K C V}{A}=K \frac{q}{A} $
$=K \sigma$