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Question

A parallel-plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.

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Solution

Initial capacitance, $C=\frac{{\in }_{0}A}{d}$ The energy of the capacitor before the insertion of the dielectric is given by ${E}_{1}=\frac{1}{2}C{V}^{2}=\frac{1}{2}\frac{{\in }_{0}A}{d}{V}^{2}$ After inserting the dielectric slab, the capacitance becomes ${C}_{1}=KC=K\frac{{\in }_{0}A}{d}$ and the final voltage becomes ${V}_{1}=\frac{V}{K}$ Thus, the final energy stored in the capacitor is given by ${E}_{2}=\frac{1}{2}\frac{K{\in }_{0}A}{d}×{\left(\frac{V}{K}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{E}_{2}=\frac{1}{2}\frac{{\in }_{0}\mathit{}A{V}^{2}}{Kd}$ Now, Work done = Change in energy ∴ W = E2 $-$ E1 $⇒W=\frac{1}{2}\frac{{\in }_{0}A{V}^{2}}{Kd}-\frac{1}{2}\frac{{\in }_{0}A{V}^{\mathit{2}}}{d}\phantom{\rule{0ex}{0ex}}⇒W=\frac{1}{2}\frac{{\in }_{0}A{V}^{2}}{d}\left(\frac{1}{K}-1\right)$

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