The potential energy of charged parallel plate capacitor is $$v_0$$. If a slab of dielectric constant $$\mathrm{K}$$ is inserted between the plates, then the new potential energy will be

Two identical parallel plate air capacitors are connected in series to a battery of emf '$$\mathrm{V}$$'. If one of the capacitor is inserted in liquid of dielectric constant '$$\mathrm{K}$$' then, potential difference of the other capacitor will become

A condenser of capacity '$$\mathrm{C}_1$$' is charged to potential '$$\mathrm{V}_1$$' and then disconnected. Uncharged capacitor of capacity '$$\mathrm{C}_2$$' is connected in parallel with '$$\mathrm{C}_1$$'. The resultant potential '$$\mathrm{V}_2$$' is

The resultant capacity between points A and B in the given circuit is