The effective capacity of the network between terminals ${A}$ and $B$ 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. $5~\text {kV}$