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):
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:
Along axis of dipole (Point Q)
The electric fields due to the positive and negative charges are:
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, (\(E_+ \gt E_-\))
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.
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