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2. Electric Potential and Capacitance
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The plates of a parallel plate capacitor of capacity $50\,\mu C$ are charged to a potential of $100\;volts$ and then separated from each other so that the distance between them is doubled. How much is the energy spent in doing so
A
$25 \times {10^{ - 2}}\,J$
B
$ - 12.5 \times {10^{ - 2}}\,J$
C
$ - 25 \times {10^{ - 2}}\,J$
D
$12.5 \times {10^{ - 2}}\,J$
Solution
(a) ${W_{ext}} = \frac{1}{2}C'V{'^2} – \frac{1}{2}C{V^2}$
$ = \left( {\frac{1}{2}} \right)\,\left( {\frac{C}{2}} \right)\,{(2V)^2} – \frac{1}{2}C{V^2} = \frac{1}{2}C{V^2}$
${W_{ext}} = \frac{1}{2} \times 50 \times {10^{ – 6}} \times {(100)^2} = 25 \times {10^{ – 2}}\,J$
Standard 12
Physics
