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 \(+2000~\text{V}\) is given to the point \(A\), the potential at \(B\) is:
             

The figure shows some of the equipotential surfaces. The magnitude and direction of the electric field are 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 \(V = 4~\text{volt}\) each plate of the capacitor has a surface area of\(10^{-2}~\text{m}^2\) and the plates are$\mathrm{}$ \(0.1\times10^{-3}~\text{m}\)apart, then the number of excess electrons on the negative plate is:

An air capacitor of capacity \(C= 10~\mu\text{F}\) 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. \(120~\mu\text{C}\)
2. \(699~\mu\text{C}\)
3. \(480~\mu\text{C}\)
4. \(24~\mu\text{C}\)

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. \(\frac{\sqrt{2} Q^{2}}{4 \pi \varepsilon_{0} a}\)
3. \(\frac{\sqrt{2} Q^{2}}{\pi \varepsilon_{0} a}\)
4. \(\frac{Q^{2}}{2 \pi \varepsilon_{0} a}\)

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 \(50~\text{V}\)?