ELECTRIC POTENTIAL AT ANY POINT DUE TO AN ELECTRIC DIPOLE :
Obtain expression for the electric potential at any point due to an electric dipole. Rewrite this expression if point of observation lies on the (i) axial line of the dipole and (ii) equatorial line of the dipole.
Consider any point P at a distance r from the centre (O) of the electric dipole AB. Let OP make an angle $\theta$ with the dipole moment $\vec{p}$. Let $r_1$ and $r_2$ be the distances of point P from -q charge and +q charge of the dipole respectively.
Step 1. Potential at point P due to -q charge is given by
$V_1 = \frac{1}{4\pi\epsilon_0} \frac{(-q)}{r_1}$
Potential at point P due to +q charge is given by
$V_2 = \frac{1}{4\pi\epsilon_0} \frac{q}{r_2}$
$\therefore$ Using principle of superposition, potential at point P due to the dipole is given by
$V = V_1 + V_2$
or
$V = -\frac{1}{4\pi\epsilon_0} \frac{q}{r_1} + \frac{1}{4\pi\epsilon_0} \frac{q}{r_2}$
$V= \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{r_2} - \frac{1}{r_1} \right]$ ...(1)
STEP 2. Draw a perpendicular from A which meets the line OP at C when produced backward. Also draw BD perpendicular on OP.
Then $r_1 = AP = CP = OP + OC$
$(\because \text{ from } \triangle AOC, OC = l\cos\theta)$
$r_1= r + l\cos\theta$
and $r_2 = BP = DP = OP - OD$
$(\because \text{ from } \triangle BOD, OD = l\cos\theta)$
$r_2= r - l\cos\theta$
STEP 3. Substituting the values of $r_1$ and $r_2$ in eqn. (1), we get,
$V = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{r - l\cos\theta} - \frac{1}{r + l\cos\theta} \right)$
$V= \frac{q}{4\pi\epsilon_0} \left( \frac{r + l\cos\theta - (r -l\cos\theta)}{(r - l\cos\theta)(r + l\cos\theta)} \right)$
$V= \frac{q}{4\pi\epsilon_0} \left( \frac{r + l\cos\theta - r + l\cos\theta}{r^2 - l^2\cos^2\theta} \right)$
$V = \frac{q}{4\pi\epsilon_0} \left( \frac{2l\cos\theta}{r^2 - l^2\cos^2\theta} \right)$
$V= \frac{q2l\cos\theta}{4\pi\epsilon_0 (r^2 - l^2\cos^2\theta)}$
Since, $q \times 2l = p$, where $p$ is dipole moment.
$\therefore V = \frac{p\cos\theta}{4\pi\epsilon_0 (r^2 - l^2\cos^2\theta)}$...(2)
STEP 4. If the point of observation P is far away from the centre of the electric dipole (i.e., $r >> l$), then eqn. (2) becomes
$V = \frac{p\cos\theta}{4\pi\epsilon_0 r^2} ...(3)$
Since $p\cos\theta = \vec{p} \cdot \hat{r}$, where $\hat{r}$ is unit vector directed along OP.
$V = \frac{\vec{p} \cdot \hat{r}}{4\pi\epsilon_0 r^2}$
$V= \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^3}$...(4)
Special Cases :
1. If point P lies on the axial line of the dipole i.e.
$\theta = 0^\circ, [so \cos 0^\circ = 1]$
then eqn. (3) becomes
$V = \frac{p}{4\pi\epsilon_0 r^2}$
2.If point P lies on the equatorial line of the dipole i.e. $\theta = 90^\circ$, then from eqn. (3), $V = 0 \quad [\because \cos 90^\circ = 0]$
Thus, electric potential due to a dipole is zero at all points on the equatorial line of the dipole.
The comparison of electric potential V due to an electric dipole with distance from the centre of the electric dipole on the axial line and the equatorial line.
