The capacitance of a spherical capacitor in air is $50 \,\mu F$. When it is immersed in oil,its capacitance becomes $110 \,\mu F$. Calculate the dielectric constant of the oil.

The space between the plates of a parallel plate capacitor is filled with a dielectric whose dielectric constant varies with distance $x$ as per the relation: $K(x) = K_0 + \lambda x$ (where $\lambda$ is a constant). The capacitance $C$ of the capacitor would be related to its vacuum capacitance $C_0$ by the relation:

$A$ capacitor has air as dielectric medium and two conducting plates of area $12 \,cm^2$ and they are $0.6 \,cm$ apart. When a slab of dielectric having area $12 \,cm^2$ and $0.6 \,cm$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \,cm$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\epsilon_0 = 8.834 \times 10^{-12} \,F/m$)

$A$ parallel plate capacitor has two layers of dielectrics as shown in the figure. This capacitor is connected across a battery. The ratio of the potential difference across the dielectric layers $(V_1/V_2)$ is:

$A$ 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.....$\%$

$A$ capacitor stores $60 \ \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \ \mu C$ flows through the battery. If the initial capacitance of the capacitor was $2 \ \mu F$,the amount of heat produced when the dielectric is inserted is ... $\mu J$.