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

The capacity of a condenser in which a dielectric of dielectric constant $5$ has been used, is $C$. If the dielectric is replaced by another with dielectric constant $20$, the capacity will become

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 $0$ to $3d$

A parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes $2C$, the dielectric constant of the medium is

A parallel plate capacitor has plates of area $A$ separated by distance $d$ between them. It is filled with a dielectric which has a dielectric constant that varies as $\mathrm{k}(\mathrm{x})=\mathrm{K}(1+\alpha \mathrm{x})$ where $\mathrm{x}$ is the distance measured from one of the plates. If $(\alpha \text {d)}<<1,$ the total capacitance of the system is best given by the expression 

A parallel plate capacitor with width $4\,cm$, length $8\,cm$ and separation between the plates of $4\,mm$ is connected to a battery of $20\,V$. A dielectric slab of dielectric constant $5$ having length $1\,cm$, width $4\,cm$ and thickness $4\,mm$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be......... $\in_{0}\,J$. (Where $\epsilon_{0}$ is the permittivity of free space)