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 \(left| {{E}_{text{equatorial}}} right|=frac{1}{2}left| {{E}_{text{axial}}} right|\)

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