A parallel plate capacitor of capacitance $$2 \mathrm{~F}$$ is charged to a potential $$\mathrm{V}$$, The energy stored in the capacitor is $$E_{1}$$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $$\mathrm{E}_{2}$$. The ratio $$\mathrm{E}_{2} / \mathrm{E}_{1}$$ is :

The distance between two plates of a capacitor is $$\mathrm{d}$$ and its capacitance is $$\mathrm{C}_{1}$$, when air is the medium between the plates. If a metal sheet of thickness $$\frac{2 d}{3}$$ and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes $$\mathrm{C}_{2}$$. The ratio $$\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$$ is

The equivalent capacitance of the combination shown is :

In this figure the resistance of the coil of galvanometer G is $$2 ~\Omega$$. The emf of the cell is $$4 \mathrm{~V}$$. The ratio of potential difference across $$\mathrm{C}_{1}$$ and $$\mathrm{C}_{2}$$ is: