In the given circuit,two identical capacitors $C_1$ and $C_2$ are connected to a battery. The space between the plates of $C_1$ is filled with air,while the space between the plates of $C_2$ is filled with a dielectric material of dielectric constant $K$. Then,

The figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plates of both capacitors is filled with a dielectric of dielectric constant $K = 3$. What will be the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric?

Assertion: When a battery remains connected,the electric potential energy increases if a dielectric material is inserted between the plates of a capacitor.
Reason: When a battery remains connected,the charge on the plates of the capacitor remains the same.

$A$ capacitor with air as the dielectric is charged to a potential of $100\;V$. If the space between the plates is now filled with a dielectric of dielectric constant $K = 10$,the potential difference between the plates will be: (in $;V$)

$A$ parallel plate capacitor has a capacity of $80 \times 10^{-6} \ F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $K = 20$. The capacitor is connected to a battery of $30 \ V$. The dielectric slab is then removed while the capacitor remains connected to the battery. Calculate the charge that passes through the wire.

$A$ parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $5$,the percentage increase in the capacitance will be.....$\%$