Two charges –q at A and +q at B separated by a small distance 2d constitute an electric dipole and its dipole moment is p (Fig.).
Let P be the point at a distance r from the midpoint of the dipole O and θ be the angle between PO and the axis of the dipole OB. Let r1 and r2 be the distances of the point P from +q and –q charges respectively.
Potential at P due to charge (+q) = q/4πε0 r1
Potential at P due to charge (−q) = -q/4πε0 r2
Total potential at P due to dipole is, V = q/4πε0 r1 + (-q/4πε0 r2)
V = q/4πε0 (1/r1- 1/r2) ...(1)
Applying cosine law,
r1^2 = r^2 + d^2 – 2rd cos θ
r1^2 = r^2 (1+ d^2/r^2 – 2d cos θ / r )
Since d is very much smaller than r, d^2/r^2 can be neglected
∴ r1 = r (1-2d cos θ / r)^1/2
or 1/r1 = 1/r (1-2d cos θ / r)^(-1/2)
Using the Binomial theorem and neglecting higher powers,
∴ 1/r1 = 1/r (1+d cos θ / r) ....(2)
Similarly,
r2^2 = r^2 + d^2 – 2rd cos (180 – θ)
or r2^2 = r^2 + d^2 + 2rd cos θ
r2 = r (1+2d cos θ / r)^1/2 ( since, d^2/r^2 is negligible)
or 1/r2 = r (1+2d cos θ / r)^(-1/2)
Using the Binomial theorem and neglecting higher powers,
∴ 1/r2 = 1/r (1-d cos θ / r) ....(3)
Substituting equation (2) and (3) in equation (1) and simplifying
V = q/4πε0 (1/r1- 1/r2)
V = q/4πε0 r (1+d cos θ / r - 1+d cos θ / r)
∴ V = 2qd cos θ / 4πε0 r^2
V = p cos θ / 4πε0 r^2 (here, p = q2d) ....(4)
Special cases :
1. When the point P lies on the axial line of the dipole on the side
of +q, then θ = 0,
∴ V = p / 4πε0 r^2
2. When the point P lies on the axial line of the dipole on the side
of –q, then θ = 180
∴ V = -p / 4πε0 r^2
3. When the point P lies on the equatorial line of the dipole, then,
θ = 90,
∴ V = 0
