If the potential of a capacitor having capacity of $6\,\mu F$ is increased from $10\, V$ to $20\, V$, then increase in its energy will be
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(b) The increase in energy of the capacitor $\Delta U = \frac{1}{2}C(V_2^2 - V_1^2) = \frac{1}{2}(6 \times {10^{ - 6}})\,({20^2} - {10^2})$
$ = 3 \times {10^{ - 6}} \times 300 = 9 \times {10^{ - 4}}\,J$
$ = 3 \times {10^{ - 6}} \times 300 = 9 \times {10^{ - 4}}\,J$
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