The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now, a dielectric of dielectric constant $K$ is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor. What will happen?

If each capacitor is of capacity $C$, find the capacitance between $A$ and $B$.

Four condensers are joined as shown in the adjoining figure. The capacity of each is $8 \mathrm{\mu F}$. The equivalent capacity between the points $A$ and $B$ will be

The capacities and connection of five capacitors are shown in the adjoining figure. The potential difference between the points $A$ and $B$ is $60 \mathrm{V}$. Then, the equivalent capacity between $A$ and $B$ and the charge on $5 \mathrm{\mu F}$ capacitance will be, respectively

In the adjoining figure, four capacitors are shown with their respective capacities and the potential difference applied. The charge and the potential difference across the $4 \mathrm{\mu F}$ capacitor will be

The resultant capacitance of the given circuit is

Three plates $A$, $B$ and $C$, each of area $50 {\mathrm{cm}}^2$, have separation $3 \mathrm{mm}$ between $A$ and $B$ and $3 \mathrm{mm}$ between $B$ and $C$. The energy stored when the plates are fully charged is

Two capacitors $A$ and $B$ are connected in series with a battery as shown in the figure. When the switch $S$ is closed and the two capacitors get charged fully, then