The electric field due to a short electric dipole at a distance $r$ on the axial line from its mid-point is $x$ times the electric field at a distance $2r$ on the equatorial line from the mid-point of the dipole. Then,the value of $x$ is

$A$ 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)

Five identical small electric dipoles are placed at a distance $r$ from an infinitely long straight wire carrying a uniform charge density $\lambda$. If they experience forces of magnitude $F_1, F_2, F_3, F_4$ and $F_5$ respectively,then:

When is the torque on an electric dipole in an electric field maximum?

Two point charges $+q$ and $-q$ are held fixed at $(-d, 0)$ and $(+d, 0)$ respectively of an $(x, y)$ coordinate system. Then:

An electric dipole consists of two particles, each of mass $1 \ kg$, separated by $1 \ m$, carrying charges $1 \ \mu C$ and $-1 \ \mu C$ respectively. It is in equilibrium in a uniform electric field of $2 \times 10^4 \ Vm^{-1}$. If it is deflected by a small angle $2^{\circ}$, the minimum time taken by it to come back again to the mean position is (in seconds): (in $\pi$)