# Explain axial field or electric field at a point in end on position, i.e., axial position of an electric dipole.

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.,

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

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.,

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

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 $$\left| \overrightarrow{{{E}_{2}}} \right|>\left| \overrightarrow{{{E}_{1}}} \right|,$$ $$E={{E}_{2}}-{{E}_{1}}=k\left[ \frac{q}{{{\left( r-a \right)}^{2}}}-\frac{q}{{{\left( r+a \right)}^{2}}} \right]$$ $$=kq\left[ \frac{{{\left( r+a \right)}^{2}}-{{\left( r-a \right)}^{2}}}{{{\left( r-a \right)}^{2}}{{\left( r+a \right)}^{2}}} \right]=kq\frac{4ra}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}$$

Or $$E=k\frac{2\times 2qa\times r}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}=k\frac{2pr}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}$$

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

$$\overrightarrow{E}=k\frac{2\overrightarrow{p}\,r}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}=\frac{1}{4\pi {{\in }_{0}}}\frac{2\overrightarrow{p}\,r}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}$$ … (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,
$$\overrightarrow{E}=k\frac{2\,\overrightarrow{p}}{{{r}^{3}}}=\frac{1}{4\pi \,{{\in }_{0}}}\frac{2\overrightarrow{p}}{{{r}^{3}}}$$ … (2)

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

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

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#### 1 Comment

1. Khlainkupar Lyngkhoi

Good