A $10\,pF$ capacitor is connected to a $50 \,V$ battery. How much electrostatic energy is stored in the capacitor
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(a) $U = \frac{1}{2}C{V^2} = \frac{1}{2} \times 10 \times {10^{ - 12}} \times {(50)^2} = 1.25 \times {10^{ - 8}}\,J$
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