A parallel plate condenser with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed, then capacitance of the capacitor becomes

Explain polarisation of polar molecule in uniform electric field.

There is an air filled $1\,pF$ parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to $2\,pF$. The dielectric constant of wax is

Eight small drops, each of radius $r$ and having same charge $q$ are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is

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 capacitor of capacitance $9 n F$ having dielectric slab of $\varepsilon_{ r }=2.4$ dielectric strength $20\, MV / m$ and $P.D. =20 \,V$ then area of plates is ....... $\times 10^{-4}\, m ^{2}$