# Electric Field Of A Dipole

## What is Electric Field of a Dipole?

A dipole is a separation of opposite electrical charges and it is quantified by an electric dipole moment. The electric dipole moment associated with two equal charges of opposite polarity separated by a distance, d is defined as the vector quantity having a magnitude equal to the product of the charge and the distance between the charges and having a direction from the negative to the positive charge along the line between the charges.

It is a useful concept in dielectrics and other applications in solid and liquid materials. These applications involve the energy of a dipole and the electric field of a dipole.

## How to Calculate Electric Field of a Dipole?

Consider an electric dipole with charges

$$\begin{array}{l}+q\end{array}$$
and
$$\begin{array}{l}–q\end{array}$$
separated by a distance of
$$\begin{array}{l}d\end{array}$$
. We shall designate components due to
$$\begin{array}{l}+q\end{array}$$
and
$$\begin{array}{l}–q\end{array}$$
using subscripts
$$\begin{array}{l}+\end{array}$$
and
$$\begin{array}{l}–\end{array}$$
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):

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

Similarly,

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

The vertical components of electric field cancel out as

$$\begin{array}{l}P\end{array}$$
is equidistant from both charges.

$$\begin{array}{l}E\end{array}$$
=
$$\begin{array}{l}E_+~cos~θ~+~E_-~cos~θ\end{array}$$

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

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

Now,

$$\begin{array}{l}cos~θ\end{array}$$
=
$$\begin{array}{l}\frac{\frac{d}{2}}{r_+}\end{array}$$
=
$$\begin{array}{l}\frac{\frac{d}{2}}{r_-}\end{array}$$
=
$$\begin{array}{l}\frac{\frac{d}{2}}{√{r^2~+~(\frac{d}{2})^2}}\end{array}$$

Substituting this value we get,

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

Dipole moment

$$\begin{array}{l}p\end{array}$$
=
$$\begin{array}{l}q~×~d\end{array}$$

When

$$\begin{array}{l}r\gt\gt d\end{array}$$
, we can neglect the
$$\begin{array}{l}\frac{d}{2}\end{array}$$
term. Thus, we have,

$$\begin{array}{l}E\end{array}$$
=
$$\begin{array}{l}\frac{1}{4πε_0} \frac{p}{r^2}^{\frac{3}{2}}\end{array}$$

$$\begin{array}{l}E\end{array}$$
=
$$\begin{array}{l}\frac{1}{4πε_0} \frac{p}{r^3}\end{array}$$

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

 $$\begin{array}{l}\overrightarrow{E}\end{array}$$ = $$\begin{array}{l}-\frac{1}{4πε_0} \frac{\overrightarrow{p}}{r^3}\end{array}$$

### Along axis of dipole (Point Q)

The electric fields due to the positive and negative charges are:

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

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

$$\begin{array}{l}E\end{array}$$
=
$$\begin{array}{l}E_+~-~E_-\end{array}$$

$$\begin{array}{l}E\end{array}$$
=
$$\begin{array}{l}\frac{1}{4πε_0} \frac{q}{(r~-~\frac{d}{2})^2}~-~\frac{1}{4πε_0}\frac{q}{(r~+~\frac{d}{2})^2}\end{array}$$

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

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

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

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

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

Factoring

$$\begin{array}{l}r^4\end{array}$$
from denominator and numerator:

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

Now if

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

$$\begin{array}{l}E\end{array}$$
=
$$\begin{array}{l}\frac{1}{4πε_0} \frac{1}{r^4} \left[\frac{2pr}{1^2}\right]\end{array}$$

$$\begin{array}{l}E\end{array}$$
=
$$\begin{array}{l}\frac{1}{4πε_0} \frac{2p}{r^3}\end{array}$$

Since in this case the electric field is along the dipole moment, (

$$\begin{array}{l}E_+ \gt E_-\end{array}$$
)

 $$\begin{array}{l}\overrightarrow{E}\end{array}$$ = $$\begin{array}{l}\frac{1}{4πε_0} \frac{2\overrightarrow{p}}{r^3}\end{array}$$

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.