A parallel plate capacitor has circular plates of 10, cm radius separated by an air-gap of 1, mm . I

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2. Electric Potential and Capacitance

medium

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

Loss of $12.5\, ergs$

Loss of $125\, ergs$

Gain of $125\, ergs$

Gain of $12.5\, ergs$

Solution

${\rm{C}} = \mathop {\frac{{{ \in _0}{\rm{A}}}}{{\rm{d}}}}\limits_{({\rm{here}}\,\,{\rm{d}} – 1{\rm{mm}})} \quad {{\rm{C}}_{{\rm{Final}}}} = \frac{{{ \in _0}{\rm{A}}}}{{10{\rm{d}}}} = \frac{{\rm{C}}}{{10}}$

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

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

$=12.5 \mathrm{\,ergs}$

Standard 12

Physics

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normal

$27$ similar drops of mercury are maintained at $10 \,V$ each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ……. times that of a smaller drop.

hard

(JEE MAIN-2021)

A $700\,pF$ capacitor is charged by a $50\,V$ battery. The electrostatic energy stored by it is

easy

In a charged capacitor, the energy resides