A capacitor stores $60\,\mu C$ charge when connected across a battery. When the gap between the plates is filled with dielectric, a charge of $120\,\mu C$ flows through the battery. The dielectric constant of the dielectric inserted is

A slab of dielectric constant $K$ has the same crosssectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}\,d$, where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be.(Given $C _{0}=$ capacitance of capacitor with air as medium between plates.)

A parallel plate capacitor has plate area $100\, m ^{2}$ and plate separation of $10\, m$. The space between the plates is filled up to a thickness $5\, m$ with a material of dielectric constant of $10 .$ The resultant capacitance of the system is $'x'$ $pF$. The value of $\varepsilon_{0}=8.85 \times 10^{-12} F \cdot m ^{-1}$ The value of $'x'$ to the nearest integer is............

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 capacitor with air as the dielectric is charged to a potential of $100\;volts$. If the space between the plates is now filled with a dielectric of dielectric constant $10$, the potential difference between the plates will be......$volts$

A parallel plate capacitor is to be designed, using a dielectric of dielectric constant $5$, so as to have a dielectric strength of $10^9\;Vm^{-1}$ . If the voltage rating of the capacitor is $12\;kV$, the minimum area of each plate required to have a capacitance of $80\;pF$ is