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

Energy per unit volume for a capacitor having area \(A\) and separation \(d\) kept at a potential difference \(V\) is given by:
1. \(\dfrac{1}{2}\varepsilon_0\dfrac{V^2}{d^2}\)
2. \(\dfrac{1}{2}\dfrac{V^2}{\varepsilon_0d^2}\)
3. \(\dfrac{1}{2}CV^2\)
4. \(\dfrac{Q^2}{2C}\)

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

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:

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:

A bullet of mass \(2~\text {gm}\) has a charge of \(2~\mu\text{C}.\) Through what potential difference must it be accelerated, starting from rest, to acquire a speed of \(10~\text{m/s}?\)
1. \(50~\text {kV}\)
2. \(5~\text {V}\)
3. \(50~\text {V}\)
4. \(5~\text {kV}\)