Two capacitors of capacitance $2C$ and $C$ are joined in parallel and charged to potential $V$. The battery is now removed and the capacitor $C$ is filled with a medium of dielectric constant $K$. The potential difference across each capacitor will be
$\frac{{3V}}{{K + 2}}$
$\frac{{3V}}{K}$
$\frac{V}{{K + 2}}$
$\frac{V}{K}$
Between the plates of a parallel plate condenser there is $1\,mm$ thick paper of dielectric constant $4$. It is charged at $100\;volt$. The electric field in $volt/metre$ between the plates of the capacitor is
A dielectric slab of dielectric constant $K$ is placed between the plates of a parallel plate capacitor carrying charge $q$. The induced charge $q^{\prime}$ on the surface of slab is given by
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in
A parallel plate capacitor, partially filled with a dielectric slab of dielectric constant $K$ , is connected with a cell of emf $V\ volt$ , as shown in the figure. Separation between the plates is $D$ . Then
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 capacitors share charges and the common potential is $V$. The dielectric constant $K$ is