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

1.	lower electric potential and lower potential energy.
2.	lower electric potential and higher potential energy.
3.	higher electric potential and lower potential energy.
4.	higher electric potential and higher potential energy.

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 electric field \(E\) between the plates with distance \(d\) as measured from plate \(P\) is correctly shown by:  
  

1.		2.
3.		4.