A parallel plate capacitor has circular plates of $10\, cm$ radius separated by an air-gap of $1\, mm$. It is charged by connecting the plates to a $100\, volt$ battery. Then the change in energy stored in the capacitor when the plates are moved to a distance of $1\, cm$ and the plates are maintained in connection with the battery, is
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${\rm{C}} = \mathop {\frac{{{ \in _0}{\rm{A}}}}{{\rm{d}}}}\limits_{({\rm{here}}\,\,{\rm{d}} - 1{\rm{mm}})} \quad {{\rm{C}}_{{\rm{Final}}}} = \frac{{{ \in _0}{\rm{A}}}}{{10{\rm{d}}}} = \frac{{\rm{C}}}{{10}}$

$\mathrm{Loss}=\mathrm{E}_{\mathrm{initial}}-\mathrm{E}_{\mathrm{final}}$

$\quad=\frac{1}{2} \mathrm{V}^{2}\left(\mathrm{C}-\frac{\mathrm{C}}{10}\right)$

$=12.5 \mathrm{\,ergs}$

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