A $700\,pF$ capacitor is charged by a $50\,V$ battery. The electrostatic energy stored by it is
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(d) $U = \frac{1}{2}C{V^2} = \frac{1}{2} \times 700 \times {10^{ - 12}}{(50)^2}$$ = 8.7 \times {10^{ - 7}}\,J$
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