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:

1. \(5.15\times 10^{9}\)
2. \(2.21\times 10^{10}\)
3. \(3.33\times 10^{9}\)
4. \(2.21\times 10^{9}\)

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

A parallel plate condenser has a capacitance \(50~\mu\text{F}\) in air and \(110~\mu\text{F}\) when immersed in an oil. The dielectric constant \(k\) of the oil is: 
1. \(0.45\)
2. \(0.55\)
3. \(1.10\)
4. \(2.20\)

Two thin dielectric slabs of dielectric constants \(K_1~\text{and}~K_2(K_{1} < K_{2})\) are inserted between plates of a parallel capacitor, as shown in the figure. The variation of the electric field \(E\) between the plates with distance \(d\) as measured from the plate \(P\) is correctly shown by:  
   

1.		2.
3.		4.

Four point charges \(-Q, -q,2q~\text{and}~2Q\) are placed, one at each corner of the square. The relation between \(Q\) and \(q\) for which the potential at the center of the square is zero, is: