Intensity of an electric field $E$ due to a dipole,depends on distance $r$ as

An electric dipole consisting of two opposite charges of $2 \times 10^{-6} \text{ C}$ separated by a distance of $3 \text{ cm}$ is placed in an electric field of $2 \times 10^5 \text{ N/C}$. The maximum torque acting on the dipole is:

The electric potential at a point due to an electric dipole is maximum and minimum when the angles between the axis of the dipole and the line joining the point to the center of the dipole are respectively:

An electric dipole of moment $p$ is placed at the origin along the $x$-axis. The electric field at a point $P$,whose position vector makes an angle $\theta$ with the $x$-axis,will make an angle $\phi$ with the $x$-axis. If $\tan \alpha = \frac{1}{2} \tan \theta$,where $\alpha$ is the angle between the electric field vector and the position vector,then the angle $\phi$ that the electric field makes with the $x$-axis is:

Two charges $+q$ and $-q$ are placed at a distance $b$ apart as shown in the figure. The electric field at a point $P$ on the perpendicular bisector as shown is

The angle between the electric field and the dipole moment at any point on the equatorial plane of an electric dipole is . . . . . . . (in $^{\circ}$)