Two capacitors each of $1\,\mu F$ capacitance are connected in parallel and are then charged by $200\;volts$ $d.c.$ supply. The total energy of their charges (in $joules$) is
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(c) $U = \frac{1}{2}C{V^2} = \frac{1}{2} \times 2 \times {(200)^2} \times {10^{ - 6}} = 0.04\,J$
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