A parallel plate capacitor has plate area '$$\mathrm{A}$$' and separation between plates is '$$d$$'. It is charged to a potential difference of $$\mathrm{V}_0$$ volt. The charging battery is then disconnected and plates are pulled apart three times the initial distance. The work done to increase the distance between the plates is $$\left(\varepsilon_0=\right.$$ permittivity of free space)

Two capacitors $$\mathrm{C}_1=3 \mu \mathrm{F}$$ and $$\mathrm{C}_2=2 \mu \mathrm{F}$$ are connected in series across d.c. source of $$100 \mathrm{~V}$$. The ratio of the potential across $$C_2$$ to $$C_1$$ is

The mean electrical energy density between plates of a charged air capacitor is (where $$\mathrm{q}=$$ charge on capacitor, $$\mathrm{A}=$$ Area of capacitor plate)

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)