A parallel plate capacitor of capacitance $12.5 \mathrm{pF}$ is charged by a battery connected between its plates to potential difference of $12.0 \mathrm{~V}$. The battery is now disconnected and a dielectric slab $\left(\epsilon_{\mathrm{r}}=6\right)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is_______.$\times 10^{-12} \mathrm{~J}$.

The distance between the plates of a parallel plate condenser is $8\,mm$ and $P.D.$ $120\;volts$. If a $6\,mm$ thick slab of dielectric constant $6$ is introduced between its plates, then

A parallel plate capacitor having crosssectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of same area but thickness $d/2$ is inserted between the plates as shown in figure having dielectric constant $K (=4) .$ The ratio of new capacitance to its original capacitance will be,

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 parallel plate capacitor having plates of area $S$ and plate separation $d$, has capacitance $C _1$ in air. When two dielectrics of different relative permittivities $\left(\varepsilon_1=2\right.$ and $\left.\varepsilon_2=4\right)$ are introduced between the two plates as shown in the figure, the capacitance becomes $C _2$. The ratio $\frac{ C _2}{ C _1}$ is

The parallel combination of two air filled parallel plate capacitors of capacitance $C$ and $nC$ is connected to a battery of voltage, $V$. When the capacitor are fully charged, the battery is removed and after that a dielectric material of dielectric constant $K$ is placed between the two plates of the first capacitor. The new potential difference of the combined system is