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 surface due to an electric dipole. Locate the points where the potential due to the dipole is zero.

Answer 1

(a) Electric dipole moment- The strength of the electric dipole is measured by the quantity of electric dipole moment. Its magnitude equal to the product of the magnitude of their charge and the distance between the two charges.
Electric dipole moment.
$p=q\times d$
It is a vector quantity
In vector form it is written as $\overrightarrow{p}=q\times \overrightarrow{d}$, where the direction of $\overrightarrow{d}$ is form negative charge to positive charge.
Electric Field of dipole at points on the equatorial plane
The magnitude of the electric field due to the two charges +q and -q are given by
$E_{+q}=\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2+a^2}..............\left(i\right)$
$E_{-q}=\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2+a^2}..............\left(i\right)$
$\therefore E_{+q}=E_{-q}$
The direction of $E_{+q}and{E}_{-q}$are shown in figure:
The components normal to the dipole axis cancel away.The components along the dipole axis add up.
$\therefore$ Total electric field
$E=-(E_{+q}+E_{-q})cos\theta \hat{p}$
$E=\frac{2qa}{4\pi {\epsilon }_0(r^2+a^2{)}^{3/2}}\hat{p}..........\left(iii\right)$
At large distance (r>>a) this reduce to
$E=-\frac{2qa}{4\pi {\epsilon }_0{r}^3}\hat{p}..........\left(iv\right)$
$\overrightarrow{p}=q\times 2\overline{a}\hat{p}$
$E=\frac{-\overrightarrow{p}}{4\pi {\epsilon }_0{r}^3}(r>>a)$
(b) Equipotential surface due to electric dipole- The potential due to the dipole is zero at the line bisecting the dipole length.