A parallel combination of two capacitors of capacities '$$2 ~\mathrm{C}$$' and '$$\mathrm{C}$$' is connected across $$5 \mathrm{~V}$$ battery. When they are fully charged, the charges and energies stored in them be '$$\mathrm{Q}_1$$', '$$Q_2$$' and '$$E_1$$', '$$E_2$$' respectively. Then $$\frac{E_1-E_2}{Q_1-Q_2}$$ in $$\mathrm{J} / \mathrm{C}$$ is (capacity is in Farad, charge in Coulomb and energy in J)

The capacitance of a parallel plate capacitor is $$2.5 ~\mu \mathrm{F}$$. When it is half filled with a dielectric as shown in figure, its capacitance becomes $$5 ~\mu \mathrm{F}$$. The dielectric constant of the dielectric is

The ratio of potential difference that must be applied across parallel and series combination of two capacitors $$C_1$$ and $$C_2$$ with their capacitance in the ratio $$1: 2$$ so that energy stored in these two cases becomes same is

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