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The potential energy of charged parallel plate capacitor is $v_0$. If a slab of dielectric constant $\mathrm{K}$ is inserted between the plates, then the new potential energy will be
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The correct answer is:
$\frac{v_0}{K}$
We know, $v_0=\frac{Q^2}{2 C}$
On inserting the slab of dielectric constant $\mathrm{k}$, the new capacitance $\mathrm{C}^{\prime}=\mathrm{KC}$
$\therefore \quad$ New potential energy $v_0^{\prime}=\frac{\mathrm{Q}^2}{2 \mathrm{C}^{\prime}}$
$v_0^1=\frac{Q^2}{2 \mathrm{KC}}=\frac{v_0}{\mathrm{~K}}$
On inserting the slab of dielectric constant $\mathrm{k}$, the new capacitance $\mathrm{C}^{\prime}=\mathrm{KC}$
$\therefore \quad$ New potential energy $v_0^{\prime}=\frac{\mathrm{Q}^2}{2 \mathrm{C}^{\prime}}$
$v_0^1=\frac{Q^2}{2 \mathrm{KC}}=\frac{v_0}{\mathrm{~K}}$
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