The potential energy of a system of two point negative charges,each of magnitude $2 \, \mu C$,placed $1 \, m$ apart in air is:

$A$ dipole having dipole moment $p$ is placed in front of a solid uncharged conducting sphere as shown in the diagram. The net potential at point $A$ lying on the surface of the sphere is :-

Two positive point charges of $10 \ \mu C$ and $12 \ \mu C$ are placed $10 \ cm$ apart in air. The work done to bring them $6 \ cm$ closer is (in $J$)

$A$ proton is moving away from an electron. Find the change in the potential energy of the system.

Three charges,each of magnitude $3 \mu C$,are placed on the vertices of an equilateral triangle of side $6 \ cm$. The net potential energy of the system will be nearly $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ SI \ unit\right]$. (in $J$)

Two equal charges $q$ of opposite sign separated by a distance $2a$ constitute an electric dipole of dipole moment $p$. If $P$ is a point at a distance $r$ from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle $\theta$ with the axis of the dipole,then the potential at $P$ is given by $(r \gg 2a)$ (where $p = 2qa$).