1.11 DIPOLE IN A UNIFORM EXTERNAL FIELD Consider a permanent dipole of dipole moment p in a uniform external field E, as shown in Fig. 1.19. (By permanent dipole, we mean that p exists irrespective of E; it has not been induced by E.) 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 FIGURE 1.19 Dipole in a each force multiplied by the arm of the couple (perpendicular uniform electric field. distance between the two antiparallel forces). Magnitude of torque = q E × 2 a sinq = 2 q a E sinq Its direction is normal to the plane of the paper, coming out of it. The magnitude of p × E is also p E sinq and its direction is normal to the paper, coming out of it. Thus, t = p × E (1.22) 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. Figure 1.20 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 FIGURE 1.20 Electric force on a paper. The comb, as we know, acquires charge through dipole: (a) E parallel to p, (b) E friction. But the paper is not charged. What then explains antiparallel to p. the attractive force? Taking the clue from the preceding 27 Reprint 2025-26 Physics 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! 1.12 CONTINUOUS CHARGE DISTRIBUTION We have so far dealt with charge configurations involving discrete charges q1, q2, ..., qn. One reason why we restricted to discrete charges is that the mathematical treatment is simpler and does not involve calculus. For many purposes, however, it is impractical to work in terms of discrete charges and we need to work with continuous charge distributions. For example, on the surface of a charged conductor, it is impractical to specify the charge distribution in terms of the locations of the microscopic charged constituents. It is more feasible to consider an area element DS (Fig. 1.21) on the surface of the conductor (which is very small on the macroscopic scale but big enough to include a very large number of electrons) and specify the charge DQ on that element. We then define a surface charge density s at the area element by ∆Q σ = (1.23) ∆S We can do this at different points on the conductor and thus arrive at a continuous function s, called the surface charge density. The surface charge density s so defined ignores the quantisation of charge and the discontinuity in charge distribution at the microscopic level*. s represents macroscopic surface charge density, which in a sense, is a smoothed out average of the microscopic charge density over an area element DS which, as said before, is large microscopically but small macroscopically. The units for s are C/m2. FIGURE 1.21 Similar considerations apply for a line charge distribution and a volume Definition of linear, charge distribution. The linear charge density l of a wire is defined by surface and volume ∆ Q charge densities. λ = (1.24) In each case, the ∆ l element (Dl, DS, DV) where Dl is a small line element of wire on the macroscopic scale that, chosen is small on however, includes a large number of microscopic charged constituents, the macroscopic and DQ is the charge contained in that line element. The units for l are scale but contains C/m. The volume charge density (sometimes simply called charge density) a very large number is defined in a similar manner: of microscopic constituents. ∆ Q ρ= (1.25) ∆ V where DQ is the charge included in the macroscopically small volume element DV that includes a large number of microscopic charged constituents. The units for r are C/m3. The notion of continuous charge distribution is similar to that we adopt for continuous mass distribution in mechanics. When we refer to 28 * At the microscopic level, charge distribution is discontinuous, because they are discrete charges separated by intervening space where there is no charge. Reprint 2025-26 Electric Charges and Fields the density of a liquid, we are referring to its macroscopic density. We regard it as a continuous fluid and ignore its discrete molecular constitution. The field due to a continuous charge distribution can be obtained in much the same way as for a system of discrete charges, Eq. (1.10). Suppose a continuous charge distribution in space has a charge density r. Choose any convenient origin O and let the position vector of any point in the charge distribution be r. The charge density r may vary from point to point, i.e., it is a function of r. Divide the charge distribution into small volume elements of size DV. The charge in a volume element DV is rDV. Now, consider any general point P (inside or outside the distribution) with position vector R (Fig. 1.21). Electric field due to the charge rDV is given by Coulomb’s law: 1 ρ ∆ V ∆ E = 2 rˆ' (1.26) 4 πε0 r' where r¢ is the distance between the charge element and P, and ˆr¢ is a unit vector in the direction from the charge element to P. By the superposition principle, the total electric field due to the charge distribution is obtained by summing over electric fields due to different volume elements: 1 ρ ∆V rˆ ' E ≅ Σ (1.27) all ∆V r' 2 4 πε0 Note that r, r¢, ˆ′r all can vary from point to point. In a strict mathematical method, we should let DV®0 and the sum then becomes an integral; but we omit that discussion here, for simplicity. In short, using Coulomb’s law and the superposition principle, electric field can be determined for any charge distribution, discrete or continuous or part discrete and part continuous.

1.14 Consider a uniform electric field E = 3 × 103 î N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

2.4 POTENTIAL DUE TO AN ELECTRIC DIPOLE As we learnt in the last 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 vector p whose magnitude is q × 2a and which points in the direction from –q to q (Fig. 2.5). 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 and p. Further, 49 Reprint 2025-26 Physics the field falls off, at large distance, not as 1/r 2 (typical of field due to a single charge) but as 1/r3. 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 1  q q  V = − (2.9)FIGURE 2.5 Quantities involved in the calculation 4 πε0  r1 r2  of potential due to a dipole. where r1 and r2 are the distances of the point P from q and –q, respectively. Now, by geometry, r12 = r 2 + a 2 − 2ar cosq r22 = r 2 + a 2 + 2ar cosq (2.10) We take r much greater than a ( r  a ) and retain terms only upto the first order in a/r 2 2  2a cosθ a 2  r1 = r 1 − + 2  r r  2  2a cosθ (2.11) ≅ r  1 − r  Similarly, 2 2  2a cosθ (2.12) r2 ≅ r 1 + r  Using the Binomial theorem and retaining terms upto the first order in a/r ; we obtain, 1 a 1  2a cos θ − 1 / 2 1   cos θ ≅ 1 − ≅ 1 + [2.13(a)] r1 r  r  r  r  1 a 1  2a cos θ − 1 / 2 1   cos θ ≅ 1 + ≅ 1 − [2.13(b)] r2 r  r  r  r  Using Eqs. (2.9) and (2.13) and p = 2qa, we get q 2 acosθ p cos θ V = = 4 πε0 r 2 4 πε0r 2 (2.14) 50 Now, p cos q = p.rˆ Reprint 2025-26 Electrostatic Potential and Capacitance where ˆr is the unit vector along the position vector OP. The electric potential of a dipole is then given by 1 p.rˆ V = 2 ; (r >> a) (2.15) 4 πε0 r 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 at the origin, Eq. (2.15) is, however, exact. From Eq. (2.15), potential on the dipole axis (q = 0, p ) is given by 1 p V = ± 2 (2.16) 4 πε0 r (Positive sign for q = 0, negative sign for q = p.) The potential in the equatorial plane (q = p/2) is zero. The important contrasting features of electric potential of a dipole from that due to a single charge are clear from Eqs. (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 and the dipole moment vector p. (It is, however, axially symmetric about p. That is, if you rotate the position vector r about p, keeping q 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.)