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A parallel plate capacitor has plates with area $A$ and separation $d$. A battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after slab is introduced, is
$K$
$\frac{1}{K}$
$\frac{A}{{{d^2}K}}$
$\frac{{{d^2}K}}{A}$
Solution
$\mathrm{U}=\mathrm{Q}^{2} / 2 \mathrm{C}$
Now, $\quad \mathrm{C}^{\prime}=\mathrm{K} \mathrm{C}$
As battery is disconnected, $\mathrm{Q}$ remains unaltered.
So,
${\mathrm{U}^{\prime}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}^{\prime}}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{KC}}} $
$\therefore $ ${\frac{\mathrm{U}}{\mathrm{U}^{\prime}}=\frac{\mathrm{Q}^{2} / 2 \mathrm{C}}{\mathrm{Q}^{2} / 2 \mathrm{KC}}=\mathrm{K}}$
