In the case of the dimensions of electric field and electric dipole moment,the power of mass is respectively:

$A$ 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. 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)

If a dipole of dipole moment $\vec{p}$ is placed in a uniform electric field $\vec{E}$,then the torque acting on it is given by:

Write the expression for the torque experienced by an electric dipole placed in a uniform electric field.

An electric dipole having each charge of magnitude $2 \mu C$ is placed in an electric field of intensity $8 \times 10^{4} \ N/C$. If the maximum torque acting on the dipole is $4 \times 10^{-3} \ N \cdot m$,the length of the dipole is: (in $mm$)

What is the angle between the electric dipole moment and the electric field strength due to it on the equatorial line (in $^o$)?