The electric field at a point due to an electric dipole,on an axis inclined at an angle $\theta$ $(< 90^{\circ})$ to the dipole axis,is perpendicular to the dipole axis,if the angle $\theta$ is

The electric field at point $A$ due to an electric dipole with dipole moment $p$ is perpendicular to the dipole moment vector $p$. The angle $\theta$ is:

An electric dipole is situated in a uniform electric field of intensity $E$. The dipole moment is $p$ and the moment of inertia is $I$. If the dipole is displaced slightly from the equilibrium position,then the angular frequency of its oscillations is:

Electrical field intensity due to an electric dipole on its axis at distance $x$ $(x \gg a)$ and on the equatorial line at distance $y$ $(y \gg a)$ are same. What is the ratio of $x$ and $y$?

Obtain the expression for the electric field due to an electric dipole at a point on its axial line.

One end of a spring of negligible unstretched length and spring constant $k$ is fixed at the origin $(0,0)$. $A$ point particle of mass $m$ carrying a positive charge $q$ is attached at its other end. The entire system is kept on a smooth horizontal surface. When a point dipole $\overrightarrow{p}$ pointing towards the charge $q$ is fixed at the origin,the spring gets stretched to a length $\ell$ and attains a new equilibrium position. If the point mass is now displaced slightly by $\Delta \ell \ll \ell$ from its equilibrium position and released,it is found to oscillate at frequency $\frac{1}{\delta} \sqrt{\frac{k}{m}}$. The value of $\delta$ is. . . . . .