 # Explain equatorial field or electric field on the equatorial line of an electric dipole i.e. broadside on position

A line perpendicular to the axial line and passing through the midpoint of electric dipole is called its equatorial line. Consider a point P on the equatorial line of an electric dipole (situated in a vacuum) at a distance r from the midpoint O of the electric dipole of length 2a as shown in figure.

Electric intensity at P due to –q charge at A, i.e.,

${{E}_{1}}=kfrac{q}{A{{P}^{2}}}=kfrac{q}{left( {{r}^{2}}+{{a}^{2}} right)}$

It is represented in magnitude and direction by $overrightarrow{PQ},$ i.e., $overrightarrow{{{E}_{1}}}.$

Electric intensity at P due to +q charge at B, i.e.,

${{E}_{2}}=kfrac{q}{B{{P}^{2}}}=kfrac{q}{left( {{r}^{2}}+{{a}^{2}} right)}$

It is represented in magnitude and direction by $overrightarrow{PR}$, i.e., $overrightarrow{{{E}_{2}}.}$

Let $overrightarrow{E}$ be the resultant electric intensity at P. According to the principle of superposition of electric fields, $overrightarrow{E}=overrightarrow{{{E}_{1}}}+overrightarrow{{{E}_{2}}}$

The resultant $left( overrightarrow{E} right),of,overrightarrow{{{E}_{1}}},and,overrightarrow{{{E}_{2}}}$ is given by the diagonal $overrightarrow{PS}$ of parallelogram PQSR.

Since ${{E}_{1}}={{E}_{2}},and,PA=BP,,overrightarrow{{{E}_{1}}},,and,,overrightarrow{{{E}_{2}}}$ can be represented by the sides $overrightarrow{PA},and,overrightarrow{BP}$. By the triangle law of vectors, the resultant electric field $left( overrightarrow{E} right),of,{{overrightarrow{E}}_{1}},and,{{overrightarrow{E}}_{2}}$ is given by the third side $overrightarrow{BA}$ of the ΔBPA (taken in the opposite order).

According to the triangle law of vectors,

$frac{{{E}_{1}}}{PA}=frac{E}{BA}$

Or $E={{E}_{1}}frac{BA}{PA}=kfrac{q}{left( {{r}^{2}}+{{a}^{2}} right)}frac{2a}{{{left( {{r}^{2}}+{{a}^{2}} right)}^{1/2}}}$

Or $E=kfrac{p}{{{left( {{r}^{2}}+{{a}^{2}} right)}^{3/2}}}$ … (1)

Since $overrightarrow{E},and,overrightarrow{p}$ are in opposite directions,

$overrightarrow{E}=-kfrac{overrightarrow{p}}{{{left( {{r}^{2}}+{{a}^{2}} right)}^{3/2}}}$ … (2)

Special cases:

1. If r is very large as compared to a (i.e., the dipole is short), a2 can be neglected as compared to r2.

Thus, from eqn (2), $overrightarrow{E}=-kfrac{overrightarrow{p}}{{{r}^{3}}}=-frac{1}{4pi {{in }_{0}}}frac{overrightarrow{p}}{{{r}^{3}}}$

1. If the dipole is situated in medium of relative permittivity ${{in }_{r}},$ then
$E=-kfrac{overrightarrow{p}}{{{in }_{r}}{{r}^{3}}}=-frac{1}{4pi {{in }_{0}}{{in }_{r}}}frac{overrightarrow{p}}{{{r}^{3}}}$
1. Short dipole,
$frac{left| {{E}_{text{equatorial}}} right|}{left| {{E}_{text{axial}}} right|}=frac{kleft( p/{{r}^{3}} right)}{kleft( 2p/{{r}^{3}} right)}=frac{1}{2}$

Or $left| {{E}_{text{equatorial}}} right|=frac{1}{2}left| {{E}_{text{axial}}} right|$