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 a direction 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.
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:
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 $\mathbf{E_{-q}}$ due to the charge –q and $\mathbf{E_{+q}}$ due to the charge q, by the parallelogram law of vectors.
(i) For points on the axis
Let the point P be at distance r from the centre of the dipole on the side of the charge q, as shown in Fig. 1(a). Then
$\mathbf{E_{-q}}=-\frac{q}{4\pi \epsilon_0 (r+a)^2}\mathbf{p}$ (1a)
where $\hat{\mathbf{p}}$ is the unit vector along the dipole axis (from –q to q). Also
$\mathbf{E_{+q}}=\frac{q}{4\pi \epsilon_0 (r-a)^2}\mathbf{p}$ (1b)
The total field at P is
$\mathbf{E}=\mathbf{E_{+q}}+\mathbf{E_{-q}}=\frac{q}{4\pi \epsilon}\left[\frac{1}{(r-a)^2}+\frac{1}{(r+a)^2} \right] \mathbf{p}$
$\mathbf{E}=\frac{q}{4\pi \epsilon}\frac{4ar}{(r^2-a^2)^2}\mathbf{p}$ (2)
For r >> a
$\mathbf{E}=\frac{q}{4\pi \epsilon}\frac{4qa}{(r^3}\hat{\mathbf{p}}$ (r >> a) (3)
(ii) For points on the equatorial plane
$E_{+q}=\frac{q}{4\pi \epsilon_0}\frac{1}{r^2+a^2}$ (4a)
$E_{-q}=\frac{q}{4\pi \epsilon_0}\frac{1}{r^2+a^2}$ (4b)
and are equal.
The directions of $\mathbf{E_{+q}}$ and $\mathbf{E_{-q}}$ are as shown in Fig. 1(b). 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 $\hat{\mathbf{p}}$ . We have
$\mathbf{E}=-(E_{+q}-E_{-q}) \ cos \theta \ \hat{\mathbf{p}}$
$\mathbf{E}=-\frac{2qa}{4\pi \epsilon_0(r^2+a^2)^{3/2}}\mathbf{p}$ (5)
At large distances (r >> a), this reduces to
$\mathbf{E}=-\frac{2qa}{4\pi \epsilon_0r^3}\mathbf{p}$ (r>>a) (6)
From Eqs. (3) and (6), it is clear that the dipole field at large distances does not involve q and a separately; it depends on the product qa. This suggests the definition of dipole moment. The dipole moment vector p of an electric dipole is defined by
$\mathbf{p}=q \times 2a \hat{\mathbf{p}}$ (7)
that is, it is a vector whose magnitude is charge q times the separation 2a (between the pair of charges q, –q) and the direction is along the line from –q to q. In terms of p, the electric field of a dipole at large distances takes simple forms:
At a point on the dipole axis
$\mathbf{E}=\frac{2\mathbf{p}}{4\pi \epsilon_0r^3}\mathbf{p}$ (r>>a) (8)
At a point on the equatorial plane
$\mathbf{E}=-\frac{2\mathbf{p}}{4\pi \epsilon_0r^3}\mathbf{p}$ (r>>a) (9)
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 and the dipole moment p.
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, Eqs. (8) and (9) are exact, true for any r.
Physical significance of dipoles
In most molecules, the centres of positive charges and of negative charges (Centre of a collection of positive point charges is defined much the same way as the centre of mass: $\mathbf{r_{cm}}=\frac{\Sigma_iq_i \mathbf{r_i}}{\Sigma_iq_i}$) lie at the same place. Therefore, their dipole moment is zero. $CO_2$ and $CH_4$ 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 of positive charges do not coincide. Therefore they have a permanent electric dipole moment, even in the absence of an electric field. Such molecules are called polar molecules. Water molecules, $H_2O$, is an example of this type. Various materials give rise to interesting properties and important applications in the presence or absence of electric field.
Example 1
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(a), and (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(b).
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| Fig.2 |
Solution:
(a) Field at P due to charge +10 μC
$=(8.987 \times 10^9 \ Nm^2/C^2)(10^{-5} \ C)\left(\frac{1}{(15-0.25)^2 \times 10^{-4} m^2} \right)$
$=4.13 \times 10^6 \ N/C$ along BP
Field at P due to charge –10 μC
$=(8.987 \times 10^9 \ Nm^2/C^2)(10^{-5} \ C)\left(\frac{1}{(15+0.25)^2 \times 10^{-4} m^2}\right)$
$=3.86 \times 10^6 \ N/C$ along PA
The resultant electric field at P due to the two charges at A and B is $=2.7 \times 10^5 \ N/C$ 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
$E=\frac{2p}{4\pi \epsilon_0r^3}$ (r/a >> 1)
where p = 2aq 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,
$p=10^{-5} \ C \times 5 \times 10^{-3} \ m=5 \times 10^{-8} \ C.m$
Therefore,
$E=(8.987 \times 10^9 \ Nm^2/C^2)(2 \times 5 \times 10^{-8} \ C.m)\left(\frac{1}{(15)^3 \times 10^{-6} m^3}\right)$
$E=2.6 \times 10^5 \ N/C$
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
$=(8.987 \times 10^9 \ Nm^2/C^2)(10^{-5} \ C)\left(\frac{1}{(15^2+0.25^2) \times 10^{-4} m^2}\right)$
$=3.99 \times 10^6 \ N/C$ along BQ
Field at Q due to charge –10 μC at A
$=(8.987 \times 10^9 \ Nm^2/C^2)(10^{-5} \ C)\left(\frac{1}{(15^2+0.25^2) \times 10^{-4} m^2}\right)$
$=3.99 \times 10^6 \ N/C$ 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
$=2 \times \frac{0.25}{\sqrt{15^2+0.25^2}}\times 3.99 \times 10^6 \ N/C$ along BA
$=1.33 \times 10^5 \ N/C$ 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:
$E=\frac{p}{4\pi \epsilon_0r^3}$ (r/a >> 1)
$E=(8.987 \times 10^9 \ Nm^2/C^2)(5 \times 10^{-8} \ C.m) \left(\frac{1}{(15)^3 \times 10^{-6} m^3}\right)$
$E=1.33 \times 10^5 \ N/C$
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.
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