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
The charge on the condenser will be doubled
The charge on the condenser will be reduced to half
The $P.D.$ across the condenser will be $320\;volts$
The $P.D.$ across the condenser will be $45\;volts$
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
An air filled parallel plate capacitor has capacity $C$. If distance between plates is doubled and it is immersed in a liquid then capacity becomes twice. Dielectric constant of the liquid is
Write the relation between $\vec P$ and $\vec E$ for a linear isotropic dielectric.
In the figure a capacitor is filled with dielectrics. The resultant capacitance 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