Q 12-02-181JEE MainJEE Main 2019 (9 Apr, Shift 1)Easy
A capacitor with capacitance $5\ \mu\text{F}$ is charged to $5\ \mu\text{C}$. If the plates are pulled apart to reduce the capacitance to $2\ \mu\text{F}$, how much work is done?
Answer: (D) $3.75\times10^{-6}\ \text{J}$
The charge stays $5\ \mu\text{C}$ (isolated capacitor). Work done equals the increase in stored energy $\dfrac{Q^2}{2C}$:
$$W = \frac{Q^2}{2}\left(\frac1{C_2} - \frac1{C_1}\right) = \frac{25\times10^{-12}}{2}\left(\frac{1}{2\times10^{-6}} - \frac{1}{5\times10^{-6}}\right)$$
$$W = 12.5\times10^{-12}\times3\times10^5 = 3.75\times10^{-6}\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics