# A capacitor with capacitance $5\; \mu F$ is charged to $5\; \mu C$. If the plates are pulled apart to reduce the capacitance to $2\; \mu F,$ how much work is done ?Option 1) $6.25\times 10^{-6}J$Option 2) $3.75\times 10^{-6}J$Option 3)$2.16\times 10^{-6}J$Option 4) $2.55\times 10^{-6}J$

$W=\; uf-ui$

$=\frac{Q^{2}}{2}\left [ \frac{1}{Cf}-\frac{1}{Ci} \right ]$

$=\left ( \frac{5\mu C}{2} \right )^{2}\left [ \frac{1}{2\mu F}-\frac{1}{5\mu F} \right ]$

$=\frac{25\times 10^{-6}}{2}\left [ \frac{3}{10} \right ]$

$=\frac{7\; 5}{2}\times 10^{-7}$

$=37.5\times 10^{-7}J$

$=3.75\times 10^{-6}J$

Option 1)

$6.25\times 10^{-6}J$

Option 2)

$3.75\times 10^{-6}J$

Option 3)

$2.16\times 10^{-6}J$

Option 4)

$2.55\times 10^{-6}J$

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