Match Column-$I$ with Column-$II$ related to an electric dipole of dipole moment $\vec{p}$ that is placed in a uniform electric field $\overrightarrow{E}$.

Column-$I$ (Angle between $\vec{p}$ and $\vec{E}$)	Column-$II$ (Potential energy of the dipole)
$a. 180^{\circ}$	$i. -pE$
$b. 120^{\circ}$	$ii. pE$
$c. 90^{\circ}$	$iii. \frac{1}{2} pE$
	$iv. 0$

The direction of the electric field at a point on the equatorial line of an electric dipole 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)

An electric dipole is placed in an electric field generated by a point charge.

$A$ wheel having mass $m$ has charges $+q$ and $-q$ on diametrically opposite points. It remains in equilibrium on a rough inclined plane in the presence of a vertical electric field $E$. Then the value of $E$ is

Two charges of $+25 \times 10^{-9} \, C$ and $-25 \times 10^{-9} \, C$ are placed $6 \, m$ apart. Find the ratio of the electric field intensity at a point $4 \, m$ from the center of the electric dipole on the $(i)$ axial line and $(ii)$ equatorial line.