Consider a permanent dipole of dipole moment p in a uniform external field E, as shown in Fig. 1. (By permanent dipole, we mean that p exists irrespective of E; it has not been induced by E.)
 |
| Fig.1: Dipole in a uniform electric field. |
There is a force qE on q and a force –qE on –q. The net force on the dipole is zero, since E is uniform. However, the charges are separated, so the forces act at different points, resulting in a torque on the dipole. When the net force is zero, the torque (couple) is independent of the origin. Its magnitude equals the magnitude of each force multiplied by the arm of the couple (perpendicular distance between the two antiparallel forces).
Magnitude of torque = qE × 2a sin θ
= 2qaE sinθ
Its direction is normal to the plane of the paper, coming out of it.
The product of the charge and the separation is the magnitude of a quantity called the electric dipole moment, denoted by :
p = qa (magnitude of electric dipole moment) (1)
The units of are charge times distance For example, the magnitude of the electric dipole moment of a water molecule is $p = 6.13 \times 10^{-30} \ C.m$
(The symbol p has multiple meanings Be careful not to confuse dipole moment with momentum or pressure. There aren’t as many letters in the alphabet as there are physical quantities, so some letters are used several times. The context usually makes it clear what we mean, but be careful.)
The magnitude of $\mathbf{p \times E}$ is also pE sin θ and its direction is normal to the paper, coming out of it. Thus,
τ $= \mathbf{p \times E}$ (2)
This torque will tend to align the dipole with the field E. When p is aligned with E, the torque is zero.
What happens if the field is not uniform? In that case, the net force will evidently be non-zero. In addition there will, in general, be a torque on the system as before. The general case is involved, so let us consider the simpler situations when p is parallel to E or antiparallel to E. In either case, the net torque is zero, but there is a net force on the dipole if E is not uniform.
 |
| Fig.2: Electric force on a dipole: (a) E parallel to p, (b) E antiparallel to p. |
Figure 2 is self-explanatory. It is easily seen that when p is parallel to E, the dipole has a net force in the direction of increasing field. When p is antiparallel to E, the net force on the dipole is in the direction of decreasing field. In general, the force depends on the orientation of p with respect to E.
This brings us to a common observation in frictional electricity. A comb run through dry hair attracts pieces of paper. The comb, as we know, acquires charge through friction. But the paper is not charged. What then explains the attractive force? Taking the clue from the preceding discussion, the charged comb ‘polarises’ the piece of paper, i.e., induces a net dipole moment in the direction of field. Further, the electric field due to the comb is not uniform. This non-uniformity of the field makes a dipole to experience a net force on it. In this situation, it is easily seen that the paper should move in the direction of the comb!
Potential Energy of an Electric Dipole
When a dipole changes direction in an electric field, the electric-field torque does work on it, with a corresponding change in potential energy. The work dW done by a torque $\tau$ during an infinitesimal displacement $d\theta$ is given by $dW=\tau d\theta$. Because the torque is in the direction of decreasing $d\theta$ we must write the torque as $\tau=-pE \ sin \ \theta$ and
$dW=\tau d\theta=-pE \ sin \ \theta d\theta$
In a finite displacement from $\theta_1$ to $\theta_2$ the total work done on the dipole is
$W=\int_{\theta_1}^{\theta_2}(-pE \ sin \ \theta )d\theta$
$W=pE \ cos \ \theta_2 - pE \ cos \ \theta_1$
The work is the negative of the change of potential energy, $W = U_2-U_1$. So a suitable definition of potential energy for this system is
$U(\theta)=pE \ cos \ \theta$
In this expression we recognize the scalar product p.E = $pE \ cos \ \theta$ so we can also write
$U(\theta)=-\mathbf{p.E}$ (potential energy for a dipole in an electric field) (3)
The potential energy has its minimum (most negative) value at the stable equilibrium position, where $\theta = 0$ and is parallel to . The potential energy is maximum when $\theta = \pi$ and is antiparallel to ; then . At $\theta = \pi/2$, where p is perpendicular to , U is zero. We could define differently so that it is zero at some other orientation of , but our definition is simplest.
Equation (3) gives us another way to look at the effect shown in Fig. 3. The electric field E gives each grass seed an electric dipole moment, and the grass seed then aligns itself with to minimize the potential energy.
 |
| Fig.3: (a)
Electric field lines produced by two equal point charges. The pattern is formed
by grass seeds floating on a liquid above two charged wires. (b) The electric field causes polarization of the grass
seeds, which in turn causes the seeds to align with the field. |
Example 1
Figure 3a, shows an electric dipole in a uniform electric field of magnitude that is directed parallel to the plane of the figure. The charges are both lie in the plane and are separated by Find (a) the net force exerted by the field on the dipole; (b) the magnitude and direction of the electric dipole moment; (c) the magnitude and direction of the torque; (d) the potential energy of the system in the position shown.
 |
| Fig.4 |
Solution:
This problem uses the ideas of this section about an electric dipole placed in an electric field. We use the relationship for each point charge to find the force on the dipole as a whole. Equation (1) gives the dipole moment, Eq. (2) gives the torque on the dipole, and Eq. (3) gives the potential energy of the system.
(a) The field is uniform, so the forces on the two charges are equal and opposite. Hence the total force on the dipole is zero.
(b) The magnitude of the electric dipole moment p is
$p=qd=(1.6 \times 10^{-19} \ C)(0.125 \times 10^{-9} \ m)$
$p=2.0 \times 10^{-29} \ C.m$
The direction of p is from the negative to the positive charge, $145^0$ clockwise from the electric-field direction (Fig. 3b).
(c) The magnitude of the torque is
$\tau = pE \ sin \ \phi$
$\tau = (2.0 \times 10^{-29} \ C.m)(5.0 \times 10^5 \ N/C) \ sin \ 145^0$
$\tau = 5.7 \times 10^{-24} \ N.m$
From the right-hand rule for vector products (see Section 1.10), the direction of the torque τ $= \mathbf{p \times E}$ is out of the page. This corresponds to a counterclockwise torque that tends to align p with E.
(d) The potential energy
$U = -pE \ cos \ \phi$
$U = -(2.0 \times 10^{-29} \ C.m)(5.0 \times 10^5 \ N/C) \ sin \ 145^0$
$\tau = 8.2 \times 10^{-24} \ N.m$
Post a Comment for "DIPOLE IN A UNIFORM EXTERNAL FIELD"