A parallel-plate capacitor initially has a capacitance of $10 \mu \mathrm{~F}$ and is charged to 100

Question

A parallel-plate capacitor initially has a capacitance of $10 \mu \mathrm{~F}$ and is charged to 100 V . If the plate separation is reduced by half while maintaining the same charge on the capacitor, what is the resulting capacitance?

Professor PC Thomas Classes · Answer

If distance is halved, $C$ doubles to $20 \mu	ext{F}$. Charge was conserved if isolated, but here not specified.