A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:

1.	increases by a factor of 4.
2.	decreases by a factor of 2.
3.	remains the same.
4.	increases by a factor of 2.

In the given circuit if point C is connected to the earth and a potential of +2000 V is given to the point A, the potential at B is:
            

The figure shows some of the equipotential surfaces. Magnitude and direction of the electric field is given by:

When a negative charge is released and moves in the electric field, it moves towards a position of:

In the given figure if $\mathrm{V}=4 v\mathrm{olt}$, each plate of the capacitor has a surface area of ${10}^{-\mathrm{2 }}{\mathrm{m}}^2$ and the plates are $0.1\times {10}^{-\mathrm{3 }}\mathrm{m}$ apart, then the number of excess electrons on the negative plate is:

1. $5.15\times {10}^9$

2. $2.21\times {10}^{10}$

3. $3.33\times {10}^9$

4. $2.21\times {10}^9$

The capacitance of a parallel plate capacitor is C. If a dielectric slab of thickness equal to one-fourth of the plate separation and dielectric constant K is inserted between the plates, then the new capacitance will be: 

An air capacitor of capacity $C=\mathrm{10 }\mu F$ is connected to a constant voltage battery of 12 V. Now the space between the plates is filled with a liquid of dielectric constant 5. The charge that flows now from battery to the capacitor is:

1. 120 $\mu C$

2. 699 $\mu C$

3. 480 $\mu C$

4. 24 $\mu C$

Four equal charges Q are placed at the four corners of a square of each side ‘a’. Work done in removing a charge – Q from its centre to infinity is:

1. 0

2. $\frac{\sqrt{2}{Q}^2}{4\pi {\epsilon }_{0}a}$

3. $\frac{\sqrt{2}{Q}^2}{\pi {\epsilon }_{0}a}$

4. $\frac{Q^2}{2\pi {\epsilon }_{0}a}$

Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle θ with the axis of the dipole, then the potential at P is given by: (r >> 2a) (Where p = 2qa) 

How much kinetic energy will be gained by an \(\alpha\text-\text{particle}\) in going from a point at \(70~\text{V}\) to another point at \(50~\text{V}\)?