Two capacitors of capacities $2 {C}$ and ${C}$ are joined in parallel and charged up to potential ${V}$. The battery is removed and the capacitor of capacity $C$ is filled completely with a medium of dielectric constant ${K}$. The potential difference across the capacitors will now be
$\frac{3 V}{K}$
$\frac{{V}}{{K}}$
$\frac{3 V}{K+2}$
$\frac{{V}}{{K}+2}$
Putting a dielectric substance between two plates of condenser, capacity, potential and potential energy respectively
Figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant $3$. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric
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
In a medium of dielectric constant $K$, the electric field is $\vec E$ . If ${ \varepsilon _0}$ is permittivity of the free space, the electric displacement vector is
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