Explain the equatorial field or electric field on the equatorial line of an electric dipole that is?

Electric dipole:

1. Two equal and opposite charges separated by a small distance are called electric dipoles.
2. The direction of the electric dipole is from negative charge to positive charge.

The magnitude of an electric dipole is,

$\mathrm{p}=\mathrm{q}\left(2\mathrm{a}\right)$

Here,

$\mathrm{q}$ is the magnitude of the charge.

$2\mathrm{a}$ is the separation between charges.

The magnitude of electric field due to a point charge at r distance is defined as,

$\mathrm{E}=\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0}$

Consider the diagram for the given situation. Two charges of magnitude Q are placed at points A and B. Point P is the equatorial point.

a is the distance from each charge to the center of the dipole.

r be the perpendicular distance to pint P from the dipole axis.

A and B are two equal and opposite charges.

$\theta$ is the angle.

$E_{+}$ is the electric field due to a positive charge at B.

$E_{-}$ is the electric field due to a negative charge at A.

$E_R$ is the net field at point P.

The magnitude of the electric field at P (broadside on position) due to positive charge is,

$\begin{array}{c}{\mathrm{E}}_{+}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{{{\mathrm{r}}_{+}}^2}\\ {\mathrm{r}}_{+}=\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2}\\ {\mathrm{E}}_{+}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{{\left(\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2}\right)}^2}\\ {\mathrm{E}}_{+}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{({\mathrm{r}}^2+{\mathrm{a}}^2)}...\left(1\right)\end{array}$

The magnitude of the electric field at P (broadside on position) due to negative charge is,

$\begin{array}{c}{\mathrm{E}}_{-}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{{{\mathrm{r}}_{-}}^2}\\ {\mathrm{r}}_{-}=\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2}\\ {\mathrm{E}}_{-}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{{\left(\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2}\right)}^2}\\ {\mathrm{E}}_{-}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{({\mathrm{r}}^2+{\mathrm{a}}^2)}...\left(2\right)\end{array}$

From $\left(1\right)$ and $\left(2\right)$, we get

${\mathrm{E}}_{+}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{{\mathrm{r}}^2+{\mathrm{a}}^2}={\mathrm{E}}_{-}=\mathrm{E}$

By using parallelogram law vector addition for two vectors having magnitudes A and B.

$\begin{array}{c}\mathrm{R}=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos\varphi }}\\ \mathrm{\varphi }=2\mathrm{\theta }(\mathrm{\varphi }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{two}\mathrm{vectors})\\ \mathrm{A}={\mathrm{E}}_{+}\\ \mathrm{B}={\mathrm{E}}_{-}\\ \mathrm{R}={\mathrm{E}}_{\mathrm{R}}\\ {\mathrm{E}}_{\mathrm{R}}=\sqrt{{\mathrm{E}}_{+}^2+{\mathrm{E}}_{-}^2+2{\mathrm{E}}_{+}{\mathrm{E}}_{-}\cos 2\mathrm{\theta }}\\ =\sqrt{\mathrm{E}+\mathrm{E}+2\mathrm{EEcos}2\mathrm{\theta }}\\ =\mathrm{E}\sqrt{1+1+2\cos 2\mathrm{\theta }}\end{array}$

By using the trigonometric identity,

$$\begin{gathered} \sqrt{1+1+2\cos 2\mathrm{\theta }}=2\mathrm{cos\theta } \\ \mathrm{cos\theta }=\frac{\mathrm{adjcent}}{\mathrm{hypotenuse}}(\mathrm{From}\mathrm{diagram}) \\ =\frac{a}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2}} \end{gathered}$$

Then,

$\begin{array}{c}{\mathrm{E}}_{\mathrm{R}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{({\mathrm{r}}^2+{\mathrm{a}}^2)}2\mathrm{cos\theta }\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{({\mathrm{r}}^2+{\mathrm{a}}^2)}\times 2\frac{\mathrm{a}}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2}}\\ {\mathrm{E}}_{\mathrm{R}}=\frac{2\mathrm{qa}}{4{\mathrm{\pi \epsilon }}_0}\times \frac{1}{{({\mathrm{r}}^2+{\mathrm{a}}^2)}^{\frac{3}{2}}}\\ \mathrm{p}=2\mathrm{qa}\\ {\mathrm{E}}_{\mathrm{R}}=\frac{\mathrm{p}}{4{\mathrm{\pi \epsilon }}_0{({\mathrm{r}}^2+{\mathrm{a}}^2)}^{\frac{3}{2}}}\end{array}$

If $\mathrm{r}>>>\mathrm{a}$ then,

$\begin{array}{c}{\mathrm{r}}^2+{\mathrm{a}}^2\approx {\mathrm{r}}^2\\ {\mathrm{E}}_{\mathrm{R}}=\frac{\mathrm{p}}{4{\mathrm{\pi \epsilon }}_0{\left({\mathrm{r}}^2\right)}^{\frac{3}{2}}}\\ {\mathrm{E}}_{\mathrm{R}}=\frac{\mathrm{p}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}\end{array}$

Therefore, the direction of the field is opposite to the direction of the electric dipole.