Consider an electric dipole consisting of charges -q and +q and of length 2a placed in uniform electric field $\stackrel{\rightharpoonup }{E}$ makng an angle $\mathrm{\theta }$ with the direction of electric field.



Force acting on charge -q at A= $-\mathrm{q}\stackrel{\rightharpoonup }{\mathrm{E}}$ (opposite to $\stackrel{\rightharpoonup }{\mathrm{E}}$ )

Force acting on charge +q at B= $\mathrm{q}\stackrel{\rightharpoonup }{\mathrm{E}}$  (along $\stackrel{\rightharpoonup }{\mathrm{E}}$)
Thus, electric dipole is under the action of two equal and unlike parallel forces giving rise to a torque on the dipole.
The magnitude of the torque is given by
 
$$\begin{gathered} \mathrm{\tau } = \mathrm{either} \mathrm{force} \times \mathrm{perpendicular} \mathrm{distance} \mathrm{between} \mathrm{the} \mathrm{two} \mathrm{forces} \\ = \mathrm{qE}\left(\mathrm{AN}\right) \\ =\mathrm{q} \mathrm{E} (2\mathrm{a} \sin \mathrm{\theta }) \\ =\mathrm{q} \left(2\mathrm{a}\right) \mathrm{E} \mathrm{sin\theta } \\ =p\mathrm{E} \mathrm{sin\theta } \end{gathered}$$

Torque experience maximum dipole when it placed perpendicular to the dirction of Electric field i.e, 

$\mathrm{when} \mathrm{\theta }=90°$

and $\sin 90° =1$

$\therefore {\mathrm{\tau }}_{\max }= p\mathrm{E}$

Torque aligns the dipole along the direction of the electric field.

In case the test charge is not vanishingly small it will produce it's own electric field. Therefore, the measured value of electric field at the observation point will be affected and will be different from the actual value of electric field at that point. Therefore we take the test charge to be negligibly small.
 

Electric dipole moment is defined as the product of either charge and the length of the electric dipole.

Electric field on equitorial line of an electric dipole:

Consider an electric dipole consisting of charges -q and +q seperated by a distance 2a. Let P be a point on equitorial line of the dipole at a distance 'r' from the centre of the dipole as shown in fig.
 


Let, $$\begin{gathered} \stackrel{\rightharpoonup }{{\mathrm{E}}_{\mathrm{A}}} \mathrm{and} {\stackrel{\rightharpoonup }{\mathrm{E}}}_{\mathrm{B}} \mathrm{be} \mathrm{the} \mathrm{Electric} \mathrm{field} \mathrm{at} \mathrm{point} \mathrm{P} \mathrm{due} \mathrm{to} \mathrm{charge} -\mathrm{q} \mathrm{at} \mathrm{point} \mathrm{A} \mathrm{and} +\mathrm{q} \mathrm{at} \mathrm{point} \mathrm{B}. \\ \mathrm{Then}, \mathrm{resultant} \mathrm{Electric} \mathrm{field} \mathrm{at} \mathrm{point} \mathrm{P} \mathrm{is} \mathrm{given} \mathrm{by} \end{gathered}$$


To find the resultant electric field intensity due to the dipole at point P, we will represent ${\stackrel{\rightharpoonup }{\mathrm{E}}}_{A }and {\stackrel{\rightharpoonup }{E}}_B$ by the two adjacent sides PL and PM of a parallelogram. Then, diagonal PN represents the resultant Electric field due to the dipole acting along Px'. The resultant electric field can also be found using the triangle law of addition of vectors.
In $$\begin{gathered} \mathbf{∆}\mathbf{ }\mathit{P}\mathit{A}\mathit{B}\mathbf{,}\mathbf{ }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{P}\mathbf{A}}\mathbf{,}\mathbf{ }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}\mathbf{P}}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}\mathbf{A}}\mathbf{ }\mathit{r}\mathit{e}\mathit{p}\mathit{r}\mathit{e}\mathit{s}\mathit{e}\mathit{n}\mathit{t}\mathbf{ }\mathbf{ }{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}}_{\mathbf{A}}\mathbf{ }\mathbf{,}\mathbf{ }{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}}_{\mathbf{B}\mathbf{ }}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}\mathbf{ }\mathit{r}\mathit{e}\mathit{s}\mathit{p}\mathit{e}\mathit{c}\mathit{t}\mathit{i}\mathit{v}\mathit{e}\mathit{l}\mathit{y}\mathbf{.} \\ \mathbf{\therefore }\mathbf{ }\mathit{B}\mathit{y}\mathbf{ }\mathit{t}\mathit{r}\mathit{i}\mathit{a}\mathit{n}\mathit{g}\mathit{l}\mathit{e}\mathbf{ }\mathit{l}\mathit{a}\mathit{w}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{a}\mathit{d}\mathit{d}\mathit{i}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{v}\mathit{e}\mathit{c}\mathit{t}\mathit{o}\mathit{r}\mathit{s}\mathbf{,} \\ \frac{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}}{\mathbf{B}\mathbf{A}}\mathbf{=}\frac{{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}}_{\mathbf{A}}}{\mathbf{P}\mathbf{A}}\mathbf{ }\mathbf{=}\frac{\left|{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}}_{\mathbf{B}}\right|}{\mathbf{B}\mathbf{P}} \\ \mathbf{\Rightarrow }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}\mathbf{ }\mathbf{=}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}}_{\mathbf{A}}\mathbf{\times }\frac{\mathbf{B}\mathbf{A}}{\mathbf{P}\mathbf{A}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{\circ }}}\mathbf{.}\frac{\mathbf{q}}{\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}\mathbf{)}}\mathbf{\times }\mathbf{ }\frac{\mathbf{2}\mathbf{a}}{\sqrt{{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{\circ }}}\mathbf{.}\frac{\mathbf{q}\mathbf{\left(}\mathbf{2}\mathbf{a}\mathbf{\right)}}{\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{)}}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \mathit{N}\mathit{o}\mathit{w}\mathbf{,}\mathbf{ }\mathbf{q}\mathbf{\left(}\mathbf{2}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathit{p}\mathbf{ }\mathit{i}\mathit{s}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{m}\mathit{a}\mathit{g}\mathit{n}\mathit{i}\mathit{t}\mathit{u}\mathit{d}\mathit{e}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{e}\mathit{l}\mathit{e}\mathit{c}\mathit{t}\mathit{r}\mathit{i}\mathit{c}\mathbf{ }\mathit{d}\mathit{i}\mathit{p}\mathit{o}\mathit{l}\mathit{e}\mathbf{ }\mathit{m}\mathit{o}\mathit{m}\mathit{e}\mathit{n}\mathit{t}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{d}\mathit{i}\mathit{p}\mathit{o}\mathit{l}\mathit{e}\mathbf{.} \\ \mathbf{\therefore }\mathbf{ }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{\circ }}}\mathbf{·}\frac{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}}{\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{)}}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \mathit{T}\mathit{h}\mathit{e}\mathbf{ }\mathit{d}\mathit{i}\mathit{r}\mathit{e}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{e}\mathit{l}\mathit{e}\mathit{c}\mathit{t}\mathit{r}\mathit{i}\mathit{c}\mathbf{ }\mathit{f}\mathit{i}\mathit{e}\mathit{l}\mathit{d}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathbf{a}\mathbf{ }\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t}\mathbf{ }\mathit{o}\mathit{n}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{e}\mathit{q}\mathit{u}\mathit{i}\mathit{t}\mathit{o}\mathit{r}\mathit{i}\mathit{a}\mathit{l}\mathbf{ }\mathit{l}\mathit{i}\mathit{n}\mathit{e}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{d}\mathit{i}\mathit{p}\mathit{o}\mathit{l}\mathit{e}\mathbf{ }\mathit{i}\mathit{s}\mathbf{ }\mathit{f}\mathit{r}\mathit{o}\mathit{m}\mathbf{ }\mathbf{ }\mathbf{+}\mathbf{q}\mathbf{ }\mathit{t}\mathit{o}\mathbf{ }\mathbf{-}\mathbf{q}\mathbf{ }\mathbf{(}\mathit{i}\mathit{n}\mathbf{ }\mathbf{a}\mathbf{ }\mathit{d}\mathit{i}\mathit{r}\mathit{e}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{o}\mathit{p}\mathit{p}\mathit{o}\mathit{s}\mathit{i}\mathit{t}\mathit{e}\mathbf{ }\mathit{t}\mathit{o}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{d}\mathit{i}\mathit{r}\mathit{e}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{e}\mathit{l}\mathit{e}\mathit{c}\mathit{t}\mathit{r}\mathit{i}\mathit{c}\mathbf{ }\mathit{d}\mathit{i}\mathit{p}\mathit{o}\mathit{l}\mathit{e}\mathbf{ }\mathit{m}\mathit{o}\mathit{m}\mathit{e}\mathit{n}\mathit{t}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{d}\mathit{i}\mathit{p}\mathit{o}\mathit{l}\mathit{e}\mathbf{)} \\ \mathbf{\therefore }\mathbf{ }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}\mathbf{=}\mathbf{-}\mathbf{ }\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{\circ }}}\mathbf{.}\mathbf{ }\frac{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}}{\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{)}}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$ 

Since the electric dipole is held in a uniform electric field no net force is acting on the dipole. 
Force due to -q is -qE and force due to +q is qE. Hence the no net force is acting on it and the translational forces is 0.

The total number of electric field lines crossing an area placed normal to the electric field is termed as electric flux.
Electric flux is a scalar quantity, its SI unit is Nm² C^–1.
Electric flux does not depend on the size and shape of object. In this case, as the charge enclosed is same there will be no net change in the electric flux coming out of the surface.