Consider an Electric Dipole with charges \(+q\) and \(–q\) separated by a distance of \(d\). We shall designate components due to \(+q\) and \(–q\) using subscripts \(+\) and \(–\) respectively.

We shall for the sake of simplicity only calculate the fields along symmetry axes, i.e. a point \(P\) along the perpendicular bisector of the dipole and a point \(Q\) along the axis of the dipole.

**Along perpendicular bisector (point \(P\)):**

The electric fields due to the positive and negative charges (Coulomb’s law):

\(E_+\) = \(\frac{1}{(4πε_0} \frac{q}{r_+^2}\) = \(\frac{1}{4πε_0} \frac{q}{(√{r^2~+~(\frac{d}{2})^2})^2}\) = \(\frac{1}{4πε_0} \frac{q}{r^2~+~(\frac{d}{2})^2}\)

Similarly,

\(E_-\) = \(\frac{1}{4πε_0}\frac{q}{r_-^2}\) = \(\frac{1}{4πε_0} \frac{q}{r^2~+~(\frac{d}{2})^2}\)

The vertical components of electric field cancel out as \(P\) is equidistant from both charges.

\(E\) = \(E_+~cos~θ~+~E_-~cos~θ\)

\(E\) = \(\frac{1}{4πε_0} \frac{q}{r^2~+~(\frac{d}{2})^2}~cos~θ~+~ \frac{1}{4πε_0} \frac{q}{r^2~+~(\frac{d}{2})^2}~cos~θ\)

\(E\) = \(\frac{1}{4πε_0} \frac{2q}{r^2~+~(\frac{d}{2})^2}~cos~θ\)

Now,

\(cos~θ\) = \(\frac{\frac{d}{2}}{r_+}\) = \(\frac{\frac{d}{2}}{r_-}\) = \(\frac{\frac{d}{2}}{√{r^2~+~(\frac{d}{2})^2}}\)

Substituting this value we get,

\(E\) = \(\frac{1}{4πε_0} \frac{2q}{r^2~+~(\frac{d}{2})^2}\frac{\frac{d}{2}}{√{r^2~+~(\frac{d}{2})^2}}\) = \(\frac{1}{4πε_0}~ \frac{qd}{(r^2~+~(\frac{d}{2})^2)^{\frac{3}{2}}}\)

Dipole moment \(p\) = \(q~×~d\)

When \(r\gt\gt d\), we can neglect the \(\frac{d}{2}\) term. Thus we have,

\(E\) = \(\frac{1}{4πε_0} \frac{p}{r^2}^{\frac{3}{2}}\)

**\(E\) = \(\frac{1}{4πε_0} \frac{p}{r^3}\)**

The dipole moment direction is defined as pointing towards the positive charge. Thus, the direction of electric field is opposite to the dipole moment:

**\(\overrightarrow{E}\) = \(-\frac{1}{4πε_0} \frac{\overrightarrow{p}}{r^3}\)**

**Along axis of dipole (point \(Q\))**

The electric fields due to the positive and negative charges are

\(E_+\) = \(\frac{1}{4πε_0} \frac{q}{r_+^2}\) = \(\frac{1}{4πε_0} \frac{q}{(r~-~\frac{d}{2})^2}\)

\(E_-\) = \(\frac{1}{4πε_0} \frac{q}{r_-^2}\) = \(\frac{1}{4πε_0} \frac{q}{r~+~\frac{d}{2})^2}\)

Since the electric fields are along the same line but opposing directions,

\(E\) = \(E_+~-~E_-\)

\(E\) = \(\frac{1}{4πε_0} \frac{q}{(r~-~\frac{d}{2})^2}~-~\frac{1}{4πε_0}\frac{q}{(r~+~\frac{d}{2})^2}\)

\(E\) = \(\frac{q}{4πε_0} \left[\frac{1}{(r~-~\frac{d}{2})^2}~-~ \frac{1}{(r~+~\frac{d}{2})^2}\right]\)

\(E\) = \(\frac{q}{4πε_0} \left[\frac{(r~+~\frac{d}{2})^2~-~(r~-~\frac{d}{2})^2}{(r^2~-~(\frac{d}{2})^2)^2}\right]\)

\(E\) = \(\frac{q}{4πε_0} \left[\frac{4r \frac{d}{2}}{(r^2~-~(\frac{d}{2})^2)^2}\right]\)

\(E\) = \(\frac{1}{4πε_0} \left[\frac{2rqd}{(r^2~-~(\frac{d}{2})^2)^2}\right]\)

\(E\) = \(\frac{1}{4πε_0}~\left[\frac{2rp}{(r^2~-~(\frac{d}{2})^2)^2}\right]\)

Factoring \(r^4\) from denominator and numerator,

\(E\) = \(\frac{1}{4πε_0} \frac{1}{r^4} \left[\frac{2pr}{(1~-~(\frac{d}{2r})^2)^2}\right]\)

Now if \(r\gt\gt d\), we can neglect the \((\frac{d}{2r})^2\) term becomes very much smaller than 1. Thus, we can neglect this term. The equation becomes:

\(E\) = \(\frac{1}{4πε_0} \frac{1}{r^4} \left[\frac{2pr}{1^2}\right]\)

**\(E\) = \(\frac{1}{4πε_0} \frac{2p}{r^3}\)**

Since in this case the electric field is along the dipole moment, (\(E_+ \gt E_-\))

**\(\overrightarrow{E}\) = \(\frac{1}{4πε_0} \frac{2\overrightarrow{p}}{r^3}\)**

Notice that in both cases the electric field tapers quickly as the inverse of the cube of the distance. Compared to a point charge which only decreases as the inverse of the square of the distance, the dipoles field decreases much faster because it contains both a positive and negative charge. If they were brought to the same point their electric fields would cancel out completely but since they have a small distance separating them, they have a feeble electric field.

To learn more, get connected to BYJU’s Classes.