A capacitance of value $4 \mathrm{\mu F}$ charged to $50 \mathrm{V}$ is connected with another capacitance of value $2 \mathrm{\mu F}$ charged to $100 \mathrm{V}$, in such a way that plates of similar charges are connected together. The total energy in multiples of ${10}^{-2} \mathrm{J}$ before joining and after joining will be:

A $3 \mathrm{\mu F}$ capacitor is charged to a potential of $300 \mathrm{V}$ and $2 \mathrm{\mu F}$ capacitor is charged to $200 \mathrm{V}$. The capacitors are then connected in parallel with plates of opposite polarities join together. What amount of charge will flow, when the plates are so connected?

Two parallel plate capacitors whose capacities are$C$ and $2C$ respectively, are joined in parallel. These are charged by $V$ potential difference. If the battery is now removed and a dielectric of dielectric constant $K$ is filled in between the plates of the capacitor $C$, then, what will be the potential difference across each capacitor?

In the given circuit if point $C$ is connected to the earth and a potential of $+2000 \mathrm{V}$ is given to point $A$, then potential at $B$ is:

The charge flowing through the cell on closing the key $k$ is equal to :

Figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right along the branch containing the capacitors. Then:

Find the potential difference across the capacitor in volts.

An air capacitor of capacity $C=10 \mathrm{\mu F}$ is connected to a constant voltage battery of $12 \mathrm{V}$. Now the space between the plates is filled with a liquid of dielectric constant $5$. The additional charge that flows now, from the battery to the capacitor is :

Minimum number of $8 \mu \mathrm{F}$ and $250 \mathrm{V}$ capacitors required to make a combination of $16 \mu \mathrm{F}$ and $1000 \mathrm{V}$ are: