The following figure shows the initial charges on the capacitors. After the switch $S$ is closed, find the charge on the capacitor $C_1$.

The following figure shows the initial charges on the capacitors. After the switch $S$ is closed, find the work done by the battery.

Two identical capacitors each of capacity $C$ are charged up to the same potential $V$. Now their oppositely charged plates are connected together, then calculate the potential of each capacitor after connection.

Two identical capacitors each of capacity $C$ are charged up to the same potential $V$. Now their oppositely charged plates are connected together, then calculate the energy loss in the form of heat.

The work done in moving a charge $q \mathrm{C}$ on the surface of a given charged conductor of potential $V \mathrm{volts}$ is,

The distance between plates of a parallel plate capacitor is $5d$. The positively charged plate is at $x=0$ and the negatively charged plate is at $x=5d$. Two slabs-one of conductor and the other of a dielectric, both of the same thickness $d$ are inserted between the plates as shown in the figure. Potential $\left(V\right)$ versus distance $x$ graph will be

A $400 \mathrm{\mu F}$ condenser is charged at a steady rate of $100 \mathrm{\mu C}$ per second. Calculate the time required to establish a potential difference of $100 \mathrm{volts}$ between its plates.

For the circuit shown in the figure, find

Final energy stored in the capacitor after the switch is closed.