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

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. . . . . .

An electric dipole is placed at an angle of $30^\circ$ with an electric field intensity of $2 \times 10^5 \, \text{NC}^{-1}$. It experiences a torque equal to $4 \, \text{Nm}$. The charge on the dipole,if the dipole length is $2 \, \text{cm}$,is:

When a test charge is brought in from infinity along the perpendicular bisector of an electric dipole,the work done is ..........

The angle between the electric dipole moment of a dipole and the electric field strength due to it on the equatorial line is (in $^{\circ}$)

$A$ 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)