Question

(a) Define electric dipole moment. Is it a scalar or a vector ? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
(b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.

Answer 1

(a) Electric dipole moment : Electric dipole moment is defined as the product of magnitude of either charges and the distance between the positive and negative charge of the electric dipole. The strength of any electric dipole is measured by the magnitude of its electric dipole moment.

Electric dipole moment, $p=q\times a$
It is a vector quantity.
In vector form it is written as $\overrightarrow{p}=q\times \overrightarrow{a}$ where the direction of $\overrightarrow{a}$ is from negative charge to positive charge.
Electric Field of dipole at points on the equatotial plane :

Let P be the point on the equatorial plane of the dipole where electric field is to be calculated. Let its distance from either charge be 'r'. The magnitude of the electric field due to the two charges +q and -q at P is given by
$E_{+q}=\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2+a^2}$ ........ (i)
$E_{-q}=\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2+a^2}$ ........ (ii)
From equations (i) and (ii),
$E_{+q}=E_{-q}=E\left(say\right)$
The direction of $E_{+q}$ and $E_{-q}$ are shown in figure.
Total electric field at P,
${\overrightarrow{E}}_p={\overrightarrow{E}}_{+q}+{\overrightarrow{E}}_{-q}$
Magnitude of E (using|| gm law of vector addition),
$E_p=\sqrt{E_{+q}^2+E_{-q}^2+2E_{+q}{E}_{-q}cos2\theta }=\sqrt{E^2+E^2+2E^{2}cos2\theta }=\sqrt{2E^2+2E^{2}cos2\theta }=\sqrt{2E^2(1+cos2\theta )}=\sqrt{2E^{2}2co{s}^2\theta }=\sqrt{4E^{2}co{s}^2\theta }{E}_p=2Ecos\theta =2\times \frac{1}{4\pi {\epsilon }_0}\frac{q}{(r^2+a^2)}\times \frac{a}{(r^2+a^2{)}^{\frac{1}{2}}}{E}_p=\frac{2aq}{4\pi {\epsilon }_0(r^2+a^2{)}^{\frac{3}{2}}}{E}_p=\frac{P}{4\pi {\epsilon }_0(r^2+a^2{)}^{\frac{3}{2}}}[\because p=2aq]vec{E}_p=\frac{\overrightarrow{p}}{4\pi {\epsilon }_0(r^2+a^2{)}^{\frac{3}{2}}}$
(b) Equipotential surface due to electric dipole :



The potential due to the dipole is zero at the line bisecting the dipole length.