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

1.	is maximum at the surface.
2.	is maximum at the centre.
3.	remains the same throughout the conductor.
4.	is maximum somewhere between the surface and the centre.

A capacitor of capacity \(C_1\) is charged up to \(V\) volt and then connected to an uncharged capacitor \(C_2\). Then final P.D. across each will be:
1. \(\frac{C_{2} V}{C_{1} + C_{2}}\)
2. \(\frac{C_{1} V}{C_{1} + C_{2}}\)
3. \(\left(1 + \frac{C_{2}}{C_{1}}\right)\)
4. \(\left(1 - \frac{C_{2}}{C_{1}} \right) V\)

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

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. \(\frac{Q^2}{2C}\)

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:

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)

A bullet of mass \(2\) g is having a charge of \(2\) µC. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of \(10\) m/s?
1. \(50\) kV
2. \(5\) V
3. \(50\) V
4. \(5\) kV

The effective capacity of the network between terminals \(\mathrm{A}\) and \(\mathrm{B}\) is: