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 onthe plate remains constant or unchanged.
AIIMS 2008, Medium
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In the given cases, $\mathrm{V}=\mathrm{V}_{0}$ (remains constant).
Energy stored inthe capacitor $\mathrm{U}=\frac{1}{2} \mathrm{CV}^{2}$
When a dielectric slab of dielectric constant
$\mathrm{K}$ is introduced between the plates of the
condenser, then $\mathrm{C} \longrightarrow \mathrm{KC}$
So energy stored will become $K$ times.
since $Q=C V,$ So $Q$ will become $K$ times
$\therefore $ Surface charge density
$\sigma^{\prime}=\frac{\mathrm{KQ}}{\mathrm{A}}=\mathrm{K} \sigma_{0}$
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