Explain axial field or electric field on the axial line of an electric dipole i.e. end on position

A line passing through the positive and negative charges of the dipole is called the axial line of the electric dipole. Consider a point P on the axial line of a dipole (situated in a vacuum) of length 2a at a distance r from the midpoint O.

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

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

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.,

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

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, \(overrightarrow{E}=overrightarrow{{{E}_{1}}}+overrightarrow{{{E}_{2}}}\)

Since \(overrightarrow{{{E}_{1}}}\) and E2 act in opposite directions, the resultant electric intensity \(left( overrightarrow{E} right)\) at P due to the dipole is represented by \(overrightarrow{PS}.\) Clearly, as \(left| overrightarrow{{{E}_{2}}} right|>left| overrightarrow{{{E}_{1}}} right|,\) \(E={{E}_{2}}-{{E}_{1}}=kleft[ frac{q}{{{left( r-a right)}^{2}}}-frac{q}{{{left( r+a right)}^{2}}} right]\) \(=kqleft[ frac{{{left( r+a right)}^{2}}-{{left( r-a right)}^{2}}}{{{left( r-a right)}^{2}}{{left( r+a right)}^{2}}} right]=kqfrac{4ra}{{{left( {{r}^{2}}-{{a}^{2}} right)}^{2}}}\)

Or \(E=kfrac{2times 2qatimes r}{{{left( {{r}^{2}}-{{a}^{2}} right)}^{2}}}=kfrac{2pr}{{{left( {{r}^{2}}-{{a}^{2}} right)}^{2}}}\)

Since \(overrightarrow{E},and,overrightarrow{p}\) are in the same direction,

\(overrightarrow{E}=kfrac{2overrightarrow{p},r}{{{left( {{r}^{2}}-{{a}^{2}} right)}^{2}}}=frac{1}{4pi {{in }_{0}}}frac{2overrightarrow{p},r}{{{left( {{r}^{2}}-{{a}^{2}} right)}^{2}}}\) … (1)

Special cases:

  1. If r is very large as compared to a, i.e., if the dipole is short, a2 can be neglected as compared to r2 and as such,
\(overrightarrow{E}=kfrac{2,overrightarrow{p}}{{{r}^{3}}}=frac{1}{4pi ,{{in }_{0}}}frac{2overrightarrow{p}}{{{r}^{3}}}\) … (2)

  1. E varies as \(1/{{r}^{3}}\) in case of an electric dipole whereas in case of monopole, \(Epropto frac{1}{r}.\)
  2. If the electric dipole is situated in a medium of relative permittivity \({{in }_{r}},\) then
\(overrightarrow{E}=kfrac{2overrightarrow{p},r}{{{in }_{r}}{{left( {{r}^{2}}-{{a}^{2}} right)}^{2}}}=frac{1}{4pi {{in }_{0}}{{in }_{r}}}frac{2overrightarrow{p},r}{{{left( {{r}^{2}}-{{a}^{2}} right)}^{2}}}\)

and \(overrightarrow{E}=kfrac{2overrightarrow{p},r}{{{in }_{r}}{{r}^{3}}}=frac{1}{4pi {{in }_{0}}{{in }_{r}}}frac{2overrightarrow{p}}{{{r}^{3}}}\) (Short dipole) … (3)

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