Write the relation between $\vec P$ and $\vec E$ for a linear isotropic dielectric.

The potential gradient at which the dielectric of a condenser just gets punctured is called

A parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its plates that covers $1 / 3$ of the area of its plates, as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C _1$. When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_1$ and the rest of the area gets charge $Q_2$. Choose the correct option/options, igonoring edge effects.

$(A)$ $\frac{E_1}{E_2}=1$ $(B)$ $\frac{E_1}{E_2}=\frac{1}{K}$ $(C)$ $\frac{Q_1}{Q_2}=\frac{3}{K}$ $(D)$ $\frac{ C }{ C _1}=\frac{2+ K }{ K }$

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......$\%$

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

Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf, $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$, then the amount of charge which will flow through the battery is (neglect internal resistance of the battery)