An electric dipole is a pair of equal and opposite point charges q and −q, separated by a distance 2a. The line connecting the two charges defines the direction of the dipole in space. By convention, the direction from −q to +q is said to be the direction of the dipole. The mid-point of locations of −q and +q is called the centre of the dipole.

Note that the direction of the electric dipole in chemistry is taken from +q to –q. Which is opposite to what we take it in physics. By Dr. T. K. Bansal.

The total charge of the electric dipole is obviously zero. This does not mean that the field of the electric dipole is zero. Since the charge q and −q are separated by some distance, the electric fields due to them, when added, do not exactly cancel out. However, at distances much larger than the separation of the two charges forming a dipole (r >> 2a), the fields due to q and −q nearly cancel out. The electric field due to a dipole therefore falls off, at large distance, faster than like 1/r^2 (the dependence on r of the field due to a single charge q). These qualitative ideas are borne out by the explicit calculation as follows:

1.11.1 The field of an electric dipole

The electric field of the pair of charges (−q and q) at any point in space can be found out from Coulomb’s law and the superposition principle. The results are simple for the following two cases:
(i) when the point is on the dipole axis, and
(ii) when it is in the equatorial plane of the dipole, i.e., on a plane perpendicular to the dipole axis through its centre.

The electric field at any general point P is obtained by adding the electric fields E−q vector due to the charge −q and E+q vector due to the charge q, by the parallelogram law of vectors.

(i) For points on the axis
Let the point P be on the axis and at distance r from the centre of the dipole on the side of the charge q, as shown in Fig. 1.20(a).

FIGURE 1.20 Electric field of a dipole at P (a) a point on the axis

Then the electric fields at Point P due to the two charges are given by:
E−q vector = −q/4πε0 (r + a)^2 × p capEquation [1.13(a)]
where p cap is the unit vector along the dipole axis (from −q to q). Also
E+q vector = q/4πε0 (r − a)^2 × p capEquation [1.13(b)]

Then The total field at P is
E vector = (E+q) vector + (E−q) vector = q/4πε0 [1/(r − a)^2 − 1/(r + a)^2] × p cap
= q/4πε0 × 4ar/(r^2 − a^2)^2 × p cap (1.14)

For r >> a
E vector = 4q a/4πε0r^3 × p capEquation (r >> a) (1.15)

(ii) For points on the equatorial plane
The magnitudes of the electric fields due to the two charges +q and −q are given by

E+q = q/4πε0 × 1/(r^2 + a^2)Equation [1.16(a)]

E−q = q/4πε0 × 1/(r^2 + a^2) Equation [1.16(b)]
and are equal in magnitude.

The directions of E+q vector and E−q vector are as shown in Fig. 1.20(b).

FIGURE 1.20 (b) Electric field of a dipole at a point on the equatorial plane of the dipole. p is the dipole moment vector of magnitude p = q × 2a and directed from −q to q.

Clearly, the components normal to the dipole axis cancel away. The components along the dipole axis add up. The total electric field is opposite to p cap. We have

E vector = − (E+q + E−q ) cosθ p cap
= −2q a/[4πε0 (r^2 + a^2)^3/2] × p cap Equation (1.17)

At large distances (r >> a), this reduces to
E vector = −2 q a/4πε0r^3 × p cap (r >> a) Equation (1.18)

From Equations. (1.15) and (1.18), it is clear that the dipole field at large distances does not involve q and a separately; it depends on the product q a. This suggests the definition of dipole moment.
The dipole moment vector p vector of an electric dipole is defined by

p vector = q × 2a p cap Equation (1.19)

that is, it is a vector whose magnitude is charge q times the separation 2a (between the pair of charges q and −q) and the direction is along the line from −q to +q.

In terms of p vector, the electric field of a dipole at large distances takes simple forms:

At a point on the dipole axis
E vector = 2p vector /4πε0r^3 (r >> a) (1.20)

At a point on the equatorial plane
E vector = −p vector /4πε0r^3 (r >> a) (1.21)

Notice the important point that the dipole field at large distances falls off not as 1/r^2 but as 1/r^3. Further, the magnitude and the direction of the dipole field depends not only on the distance r but also on the angle between the position vector r vector and the dipole moment p vector.

We can think of the limit when the dipole size 2a approaches zero, the charge q approaches infinity in such a way that the product p = q × 2a is finite. Such a dipole is referred to as a point dipole. For a point dipole, Equations (1.20) and (1.21) are exact, true for any r.

1.11.2 Physical significance of dipoles

In most molecules, the centres of positive charges and that of negative charges* lie at the same place. Therefore, their dipole moment is zero. cO2 and CH4 are of this type of molecules. However, they develop a dipole moment when an electric field is applied. But in some molecules, the centres of negative charges and that of positive charges do not coincide. Therefore they have a permanent electric dipole moment associated with them, even in the absence of an electric field. Such molecules are called polar molecules. Water molecules, H2O, is an example of this type. Various materials give rise to interesting properties and important applications in the presence or absence of electric field.
(* Centre of a collection of positive point charges is defined much the same way as the centre of mass:

rcm vector = Σi (qI ri) / Σi (qi))

Example 1.10

Two charges ±10 μC are placed 5.0 mm apart. Determine the electric field at
(a) a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge, as shown in Fig. 1.21(a), and

Fig. 1.21(a)

(b) a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in Fig. 1.21(b).

Fig. 1.21(b)

Solution:
(a) Field at P due to charge +10 μC
= 10^−5C/4π(8.854 × 10^−12 C^2 N^−1 m^−2) × 1/(15 − 0.25)^2 × 10^−4 m^2
= 4.13 × 10^6 NC^−1 along BP

Field at P due to charge −10 μC
= 10^−5C/4π(8.854 × 10^−12 C^2 N^−1 m^−2) × 1/(15 + 0.25)^2 × 10^−4 m^2
= 3.86 × 10^6 NC^−1 along PA

The resultant electric field at P due to the two charges at A and B is
= 2.7 × 10^5 NC^−1 along BP

In this example, the ratio OP/OB is quite large (= 60). Thus, we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole. For a dipole consisting of charges ± q, 2a distance apart, the electric field at a distance r from the centre on the axis of the dipole has a magnitude

E = 2p/4πε0r^3 (r/a >> 1)
where p = 2a q is the magnitude of the dipole moment.

The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from −q to q). Here,
p =10^−5 C × 5 × 10^−3 m = 5 × 10^−8 C m

Therefore,
E = 2 × 5 × 10^−8Cm/4π(8.854 × 10^−12 C^2 N^−1 m^−2) × 1/(15)^3 × 10^−6 m^3
= 2.6 × 10^5 NC^−1

along the dipole moment direction AB, which is close to the result obtained earlier.

(b) Field at Q due to charge + 10 μC at B
= 10^−5C/4π(8.854 × 10^−12 C^2 N^−1 m^−2) × 1/[15^2 + (0.25)^2] × 10^−4 m^2
= 3.99 × 10^6 NC^−1 along BQ

Field at Q due to charge −10 μC at A
= 10^−5C/4π(8.854 × 10^−12 C^2 N^−1 m^−2) × 1/[15^2 + (0.25)^2] × 10^−4 m^2
= 3.99 × 10^6 NC^−1 along QA

Clearly, the components of these two forces with equal magnitudes cancel along the direction oQ but add up along the direction parallel to BA. Therefore, the resultant electric field at Q due to the two charges at A and B is
= 2 × 0.25/√[15^2 + (0.25)^2] × 3.99 × 10^6 NC^−1 along BA
= 1.33 × 10^5 NC^−1 along BA

As in (a), we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to the axis of the dipole:
E = p/4πε0r^3 (r/a >> 1)
E = 5 × 10^−8Cm/4π(8.854 × 10^−12 C^2 N^−1 m^−2) × 1/(15)^3 × 10^−6 m^3
= 1.33 × 10^5 NC^−1

The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again, the result agrees with that obtained before.