$A$ capacitor of capacitance $15 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.5$,dielectric strength $30 \, MV/m$,and potential difference $V = 30 \, V$. The area of the plate is ....... $\times 10^{-4} \, m^{2}$.

Two identical capacitors $1$ and $2$ are connected in series to a battery as shown in the figure. Capacitor $2$ contains a dielectric slab of dielectric constant $k$ as shown. $Q_1$ and $Q_2$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q'_1$ and $Q'_2$. Then:

Two capacitors of capacitance $2C$ and $C$ are joined in parallel and charged to potential $V$. The battery is now removed and the capacitor $C$ is filled with a medium of dielectric constant $K$. The potential difference across each capacitor will be

Explain the effect of a dielectric on the capacitance of a parallel plate capacitor and obtain the formula for the dielectric constant.

$A$ dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3d$. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from $x = 0$ to $x = 3d$:

$A$ parallel plate capacitor with air between the plates has a capacitance of $15 \, pF$. The separation between the plates is doubled and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $\frac{x}{4} \, pF$. The value of $x$ is $............$