# 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 $$+q$$ 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.