Two identical metal plates are given charges $q_1$ and $q_2\left(q_2 < q_1\right)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance ' C ', the potential difference ' $V$ ' between the plates is

Five capacitors, each of capacity ' $C$ ' are connected as shown in the figure. The resultant capacity between A and B is $14 \mu \mathrm{~F}$. The capacity of each capacitor is

Two capacitors of $100 \mu \mathrm{~F}$ and $50 \mu \mathrm{~F}$ are connected in parallel. If the potential difference across $100 \mu \mathrm{~F}$ is 20 V and across $50 \mu \mathrm{~F}$ is 40 V , then the common potential of the parallel combination will be (same polarities of the capacitor connected together)

A $4 \mu \mathrm{~F}$ capacitor is charged to 10 V . The battery is then disconnected and a pure 10 mH coil is connected across the capacitor so that LC oscillations are set up. The maximum current in the coil is