Question

derive an expression for the electric field at a point on the axial/equatorial position of an electric Dipole?

Answer 1

Hey, refer to the Image attached.

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Answer 2

The electric field at a point on the axial position is $E=\frac{2 p}{4 \pi \varepsilon_{0}}\left[\frac{1}{r^{3}}\right]$ and on equatorial position is $E=\frac{p}{4 \pi \varepsilon_{0}}\left[\frac{1}{r^{3}}\right]$.

Explanation:

Here, p is the electric dipole moment, r is the distance of the point from the midpoint of dipole.

Let two charges +q and –q of equal magnitude be placed at a distance of 2d thus forming an electric dipole between them. Let P be a point along the axial line of the dipole at a distance r from the midpoint of the dipole.

So the distance between P and +q is (r-d) and the distance between P and –q is (r+d)

The formula for electric field at a point is,

$E=k \frac{q}{r^{2}}$

Where k is constant of proportionality, q is the charge and r is the distance between the charge and the point.

$k=\frac{1}{4 \pi \varepsilon_{0}}$

So the electric field acting at point P due to the charge +q is  

$E_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(r-d)^{2}} \rightarrow(1)$

As the charge is positive, the electric field will be acting outward to the charge +q.  

And the electric field acting at point P due to the charge –q is

$E_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(r+d)^{2}} \rightarrow(2)$

As the charge is negative, the electric field will be acting inward i.e, towards the charge –q.

Thus, E_1 is acting in opposite direction to E_2.  

The net electric field acting at point P due to both the charges can be obtained by

$E=E_{1}+\left(-E_{2}\right)$

$E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(r-d)^{2}}+\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(r+d)^{2}}$

$E=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{(r-d)^{2}}+\frac{1}{(r+d)^{2}}\right]$

$E=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{(r+d)^{2}+(r-d)^{2}}{(r-d)^{2}(r+d)^{2}}\right]$

So using the basic algebra formula of (a+b)2 and (a-b)2 the above equation can be solved as

$E=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{4 r d}{\left(r^{2}-d^{2}\right)^{2}}\right]$

If P is far away from the dipole d <<< r so d can be neglected, so the above equation get reduced to  

$E=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{4 r d}{r^{4}}\right]$

The dipole moment is written as,

$p = 2qd$

Thus,

$E=\frac{4 q d}{4 \pi \varepsilon_{0}}\left[\frac{1}{r^{3}}\right]$

The electric field along the axial point is,

$E=\frac{2 p}{4 \pi \varepsilon_{0}}\left[\frac{1}{r^{3}}\right]$

If the point P lies in the equatorial points at a distance r from the midpoint of the dipole in perpendicular direction. The distance between P and the charge +q and distance between point P and charge –q will be same, i.e. $r^{2}+d^{2}$

So in this case, the magnitude of electric field due to charge $+q \ (E_1)$ will be equal to the magnitude of electric field due to charge $-q \ (E_2)$. As it is in equatorial point, the electric field of $E_1\  \text{and} \ E_2$ will be resolved in vertical and horizontal direction separately. The vertical components of both the electric fields will be equal in magnitude but acting in opposite direction, thus canceling each other.

So the net electric field at the equatorial point P is the sum of horizontal component of electric field acting on P due to charge +q and –q.

$E=E_{1} \cos \theta+E_{2} \cos \theta$

As $\mathrm{E}_{1}=\mathrm{E}_{2}$

Then,

$\text{Net electric field}E=2 E_{1} \cos \theta$

$E=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{\left(r^{2}+d^{2}\right)}\right] \times(2 \cos \theta)$

Here,

$\cos \theta=\frac{d}{\sqrt{\left(r^{2}+d^{2}\right)}}$

So,

$E=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{\left(r^{2}+d^{2}\right)}\right] \times 2 \times \frac{d}{\sqrt{\left(r^{2}+d^{2}\right)}}$

$E=\frac{p}{4 \pi \varepsilon_{0}}\left[\frac{1}{\left(r^{2}+d^{2}\right)^{3 / 2}}\right]$

For a dipole, d <<<< r  

Thus,

$E=\frac{p}{4 \pi \varepsilon_{0}}\left[\frac{1}{r^{3}}\right]$