A parallel plate capacitor of capacitance 12.5 pF is charged by a… - Vidyadip

Question

A parallel plate capacitor of capacitance $12.5 pF$ is charged by a battery connected between its plates to potential difference of $12.0 V.$ The battery is now disconnected and a dielectric slab$\left(\epsilon_{ r }=6\right)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is $..........\times 10^{-12} J$.

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Answer

Before inserting dielectric capacitance is given $C _0=12.5 pF$ and charge on the capacitor $Q = C _0 V$ After inserting dielectric capacitance will become $\in_{ r } C _0$.
Change in potential energy of the capacitor $=E_{i}-E_{f}$
$=\frac{Q^2}{2 C_{i}}-\frac{Q^2}{2 C_{f}}=\frac{Q^2}{2 C_0}\left[1-\frac{1}{\epsilon_{r}}\right]$
$=\frac{\left(C_0 V\right)^2}{2 C_0}\left[1-\frac{1}{\epsilon_{r}}\right]$
$=\frac{1}{2} C_0 V^2\left[1-\frac{1}{\epsilon_{r}}\right]$
Using $C _0=12.5 pF , V =12 V, \epsilon_{ r }=6$
$=\frac{1}{2}(12.5) \times 12^2\left[1-\frac{1}{6}\right]$
$=\frac{1}{2}(12.5) \times 12^2 \times \frac{5}{6}$
$=750 pJ=750 \times 10^{-12} J$

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