What is the difference between the electric field of a point charge and the electric field of an electric dipole?

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)

Charges are placed at the vertices of an equilateral triangle as shown in the figure. What is the resultant dipole moment?

Write the equation for the electric field produced by an electric dipole at a point on its equatorial line,given the condition $r >> a$ (where $r$ is the distance from the center and $2a$ is the dipole length).

Four charges are placed at a distance $a$ from the origin as shown in the figure. The dipole moment of the configuration is: