2.4 Potential due to an Electric Dipole

As we learnt in the previous chapter, an electric dipole consists of two charges +q and −q separated by a (small) distance 2a. Its total charge is zero. It is characterised by a dipole moment p vector whose magnitude is q × 2a and which points in the direction from −q to +q (Fig.2.5).

FIGURE 2.5: Quantities involved in the calculation of potential due to a dipole.

We also saw that the electric field of a dipole at a point with position vector r depends not just on the magnitude r, but also on the angle between r vector and p vector. Further, the field falls off, at large distances, not as 1/r^2 (typical of field due to a single point charge) but as 1/r^3. We, now, determine the electric potential due to a dipole and contrast it with the potential due to a single charge.

As before, we take the origin at the centre of the dipole. Now we know that the electric field obeys the superposition principle. Since potential is related to the work done by the field, electrostatic potential also follows the superposition principle. Thus, the potential due to the dipole is the sum of potentials due to the charges +q and −q

V = 1/4πε0 [q/r1 − q/r2] .. .. (2.9)

where r1 and r2 are the distances of point P from +q and −q respectively.

Now, by geometry,


r1^2 = r^2 + a^2 − 2 a r cos θ


r2^2 = r^2 + a^2 + 2 a r cosθ .. .. (2.10)

We take r much greater than a, (i.e. r >> a) and retain terms only upto the first order in a/r

r1^2 = r^2 [1 − (2a/r) cosθ + a^2/r^2)

≅ r^2[1 − (2 a/r) cos θ] .. .. (2.11)

Similarly,


r2^2 ≅ r^2 [1 + (2 a/r) cos θ] .. .. (2.12)

Using the Binomial theorem and retaining terms upto the first order in a/r; we obtain,

1/r1≅ 1/r [1 − (2a cos θ)/r] ^ (−1/2)


≅ 1/r [1 + a/r cos θ] .. .. (2.13 a)

1/r2 ≅ 1/r [1+ (2a cos θ)/r] ^ (−1/2)


≅ 1/r [1 − a/r cos θ] .. .. (2.13 b)

Using equations. (2.9) and (2.13) and p = 2q a, we get

V = (q/4πε0) × (2 a cos θ/r^2)

Or V = p cos θ/ (4πε0 r^2) .. .. (2.14)

Now, p cos θ = p vector ⋅ r cap

where r cap is the unit vector along the position vector oP.

The electric potential of a dipole is then given by


V = (1/4πε0) × p vector ⋅ r cap/r^2; for (r >> a) .. .. (2.15)

Equation (2.15) is, as indicated, approximately true only for distances large compared to the size of the dipole, so that higher order terms in a/r are negligible. For a point dipole p vector at the origin, Equation (2.15) is, however, exact.

From Equation (2.15), potential on the dipole axis (θ = 0, π) is given by


V axial = ± 1/4πε0 × p /r^2 .. .. (2.16)


(Positive sign for θ = 0, negative sign for θ = π).

The potential in the equatorial plane (θ = π/2) is zero.

The important contrasting features of electric potential of a dipole from that due to a single charge are clear from Equations. (2.8) and (2.15):


(i) The potential due to a dipole depends not just on r but also on the angle between the position vector r vector and the dipole moment vector p vector. (It is, however, axially symmetric about p vector. That is, if you rotate the position vector r vector about p vector, keeping θ fixed, the points corresponding to P on the cone so generated will have the same potential as at P.)


(ii) The electric dipole potential falls off, at large distance, as 1/r^2, not as 1/r, characteristic of the potential due to a single charge. (You can refer to the Fig.2.5 for graphs of 1/r^2 versus r and 1/r versus r, drawn there in another context.)