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 $45\;volts$
If a slab of insulating material $4 \times {10^{ - 3}}\,m$ thick is introduced between the plates of a parallel plate capacitor, the separation between plates has to be increased by $3.5 \times {10^{ - 3}}\,m$ to restore the capacity to original value. The dielectric constant of the material will be
Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf, $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$, then the amount of charge which will flow through the battery is (neglect internal resistance of the battery)
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.......$\mu J$
The capacity of a parallel plate condenser is $10\,\mu F$ without dielectric. Dielectric of constant $2$ is used to fill half the distance between the plates, the new capacitance in $\mu F$ is
In a parallel plate condenser, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. There is a glass slab of $3\,mm$ thick and of radius $12\,cm$ with dielectric constant $6$ between its plates. The capacity of the condenser will be