A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in
Reduction of charge on the plates and increase of potential difference across the plates
Increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates.
Decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates
None of the above
A capacitor is connected to a battery of voltage $V$. Now a di electric slab of dielectric constant $k$ is completely inserted between the plates, then the final charge on the capacitor will be
(If initial charge is $q_{0}$ )
The capacitance of a parallel plate capacitor with air as medium is $6\, \mu F$. With the introduction of a dielectric medium, the capacitance becomes $30\, \mu F$. The permittivity of the medium is..........$C ^{2} N ^{-1} m ^{-2}$
$\left(\varepsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}\right)$
The capacitance of an air filled parallel plate capacitor is $10\,p F$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $40 \times {10^{ - 12}}farads$. The dielectric constant of wax is
An air capacitor of capacity $C = 10\,\mu F$ is connected to a constant voltage battery of $12\,V$. Now the space between the plates is filled with a liquid of dielectric constant $5$. The charge that flows now from battery to the capacitor is......$\mu C$
An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel capacitor charged to a potential $V$. The two share the charge and the common potential is $V'$. The dielectric constant $K$ is