A parallel plate capacitor has a uniform electric field \(\vec{E}\) in the space between the plates. If the distance between the plates is \(d\) and the area of each plate is \(A\) the energy stored in the capacitor is: 
\(\left ( \varepsilon_{0} = \text{permittivity of free space} \right )\)
1. \(\frac{1}{2}\varepsilon_0 E^2 Ad\)
2. \(\frac{E^2 Ad}{\varepsilon_0}\)
3. \(\frac{1}{2}\varepsilon_0 E^2 \)
4. \(\varepsilon_0 EAd\)

Two charged spherical conductors of radii \(R_1\) and \(R_2\) are connected by a wire. The ratio of surface charge densities of spheres \(\left ( \frac{\sigma _{1}}{\sigma _{2}}\right )\) is:
1. \(\sqrt{\frac{R_1}{R_2}}\)
2. \(\frac{R^2_1}{R^2_2}\)
3. \(\frac{R_1}{R_2}\)
4. \(\frac{R_2}{R_1}\)

Twenty seven drops of same size are charged at \(220~\text{V}\) each. They combine to form a bigger drop. Calculate the potential of the bigger drop.
1. \(1520~\text{V}\)
2. \(1980~\text{V}\)
3. \(660~\text{V}\)
4. \(1320~\text{V}\)

The equivalent capacitance of the combination shown in the figure is: 

Three capacitors each of capacity \(4\) µF are to be connected in such a way that the effective capacitance is \(6\) µF. This can be done by:

Energy per unit volume for a capacitor having area \(A\) and separation \(d\) kept at a potential difference \(V\) is given by:
1. \(\frac{1}{2}\varepsilon_0\frac{V^2}{d^2}\)
2. \(\frac{1}{2}\frac{V^2}{\varepsilon_0d^2}\)
3. \(\frac{1}{2}CV^2\)
4. \(\frac{Q^2}{2C}\)

If identical charges (–q) are placed at each corner of a cube of side 'b' then the electrical potential energy of charge (+q) which is placed at centre of the cube will be:

1. $\frac{-4\sqrt{2}{\mathrm{q}}^2}{{\mathrm{\pi \epsilon }}_0\mathrm{b}}$

2. $\frac{-8\sqrt{2}{\mathrm{q}}^2}{{\mathrm{\pi \epsilon }}_0\mathrm{b}}$

3. $\frac{-4{\mathrm{q}}^2}{\sqrt{3}{\mathrm{\pi \epsilon }}_0\mathrm{b}}$

4. $\frac{8\sqrt{2}{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{b}}$

A capacitor of capacity C₁ is charged up to V volt and then connected to an uncharged capacitor C₂. Then final P.D. across each will be:

1. $\frac{{\mathrm{C}}_2\mathrm{V}}{{\mathrm{C}}_1+{\mathrm{C}}_2}$

2. $\frac{{\mathrm{C}}_1\mathrm{V}}{{\mathrm{C}}_1+{\mathrm{C}}_2}$

3. $\left(1+\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1}\right)$

4. $(1-\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1})\mathrm{V}$

Some charge is being given to a conductor. Then it's potential:

The energy and capacity of a charged parallel plate capacitor are \(E\) and \(C\) respectively. If a dielectric slab of ${}_{}$\(E_r=6\) is inserted in it, then the energy and capacity become:
(Assuming the charge on plates remains constant)