Consider a parallel plate capacitor of $10\,\mu F$ with air filled in the gap between the plates. Now,one half of the space between the plates is filled with a dielectric of dielectric constant $K = 4$,as shown in the figure. The capacity of the capacitor changes to.......$\mu F$.

$A$ capacitor of capacitance $9 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.4$,dielectric strength $20 \, MV/m$,and potential difference $V = 20 \, V$. The area of the plates is ....... $\times 10^{-4} \, m^{2}$.

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 the voltage source remaining connected to the capacitor.

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

Two capacitors $C$ and $2C$ are connected in parallel and charged to a potential $V$ by a battery. The battery is then removed. The capacitor $C$ is filled with a dielectric of constant $K$. What is the new potential difference across the capacitors?

Consider the arrangement shown in the figure. The total energy stored is $U_1$ when the key $K$ is closed. Now,the key $K$ is opened,and two dielectric slabs of relative permittivity $\epsilon_r$ are introduced between the plates of the two capacitors. The slabs fit tightly between the plates. The total energy stored is now $U_2$. Then,the ratio $U_1/U_2$ is: