A point charge $q$ of mass $m$ is suspended vertically by a string of length $l$. A point dipole of dipole moment $\overrightarrow{ p }$ is now brought towards $q$ from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is $N \times( mgh )$, where $g$ is the acceleration due to gravity, then the value of $N$ is. . . . . . (Note that for three coplanar forces keeping a point mass in equilibrium, $\frac{F}{\sin \theta}$ is the same for all forces, where $F$ is any one of the forces and $\theta$ is the angle between the other two forces)

Electric field at a place is $\overrightarrow {E\,}  = {E_0}\hat i\,V/m$ . A particle of charge $+q_0$  moves from point $A$ to $B$ along a circular path find work done in this motion by electric field 

Charges $-q,\, q,\,q$ are placed at the vertices $A$, $B$, $C$ respectively of an equilateral triangle of side $'a'$ as shown in the figure. If charge $-q$ is released keeping remaining two charges fixed, then the kinetic energy of charge $(-q)$ at the instant when it passes through the mid point $M$ of side $BC$ is 

The diagram shows three infinitely long uniform line charges placed on the $X, Y $ and $Z$ axis. The work done in moving a unit positive charge from $(1, 1, 1) $ to $(0, 1, 1) $ is equal to

In the figure shown the electric potential energy of the system is: ( $q$ is at the centre of the conducting neutral spherical shell of inner radius $a$ and outer radius $b$ )