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Keeping the voltage of the charging source constant, what would be the percentage change in the energy stored in a parallel plate capacitor, if the separation between its plates were to be decreased by $10 \% $?

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Energy stored in a parallel plate capacitor is $ U=\large\frac{1}{2}$$CV^2$
$\qquad =\large\frac{1}{2} \frac{\in_0 A}{d}$$V^2$------(i)
When separation between the plates is decreased by 10%, $d'=d \large\frac{10}{100}$$d=0.9 d$
Change in energy $(U' - U)=\large\frac{1}{2} \in_0 $$AV^2 \bigg[ \large\frac{1}{0.9} -\frac{1}{1}\bigg]$
% age change in energy= $\large\frac{(U'-U)}{U} $$ \times 100 =\large\frac{100}{9} $$=11.1 \%$
answered Jun 19, 2014 by meena.p

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