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.)

2.5 POTENTIAL DUE TO A SYSTEM OF CHARGES Consider a system of charges q1, q2,…, qn with position vectors r1, r2,…, rn relative to some origin (Fig. 2.6). The potential V1 at P due to the charge q1 is 1 q1 V1 = 4 πε0 r1P where r1P is the distance between q1 and P. Similarly, the potential V2 at P due to q2 and V3 due to q3 are given by 1 q 2 1 q 3 V 2 = , V 3 = 4 πε0 r2P 4 πε0 r3P where r2P and r3P are the distances of P from charges q2 and q3, respectively; and so on for the potential due to other charges. By the FIGURE 2.6 Potential at a point due to a superposition principle, the potential V at P due system of charges is the sum of potentials to the total charge configuration is the algebraic due to individual charges. sum of the potentials due to the individual charges V = V1 + V2 + ... + Vn (2.17) 51 Reprint 2025-26 Physics 1  q1 q 2 q n  = + + ...... + (2.18) 4 πε0  r1P r2 P rnP  If we have a continuous charge distribution characterised by a charge density r (r), we divide it, as before, into small volume elements each of size Dv and carrying a charge rDv. We then determine the potential due to each volume element and sum (strictly speaking , integrate) over all such contributions, and thus determine the potential due to the entire distribution. We have seen in Chapter 1 that for a uniformly charged spherical shell, the electric field outside the shell is as if the entire charge is concentrated at the centre. Thus, the potential outside the shell is given by 1 q V = (r ≥ R ) [2.19(a)] 4 πε0 r where q is the total charge on the shell and R its radius. The electric field inside the shell is zero. This implies (Section 2.6) that potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface, which is 1 q V = [2.19(b)] 4 πε0 R Example 2.2 Two charges 3 × 10–8 C and –2 × 10–8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. Solution Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin (Fig. 2.7). FIGURE 2.7 Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have 1  3 × 10 – 8 2 × 10 –8  − x × 10 –2 4 πε0  (15 − x ) × 10 –2  = 0 where x is in cm. That is, 3 2 − = 0 2.2 x 15 − x which gives x = 9 cm. If x lies on the extended line OA, the required condition is 3 2 − = 0 EXAMPLE x x − 15 Reprint 2025-26 Electrostatic Potential and Capacitance which gives x = 45 cm Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the EXAMPLE formula for potential used in the calculation required choosing potential to be zero at infinity. 2.2 Example 2.3 Figures 2.8 (a) and (b) show the field lines of a positive and negative point charge respectively. Electric potential, equipotential-sufaces-12584/ FIGURE 2.8 equipotential (a) Give the signs of the potential difference VP – VQ; VB – VA. (b) Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B. surfaces: (c) Give the sign of the work done by the field in moving a small positive charge from Q to P. (d) Give the sign of the work done by the external agency in moving a small negative charge from B to A. (e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A? Solution 1 (a) As V ∝ , VP > VQ. Thus, (VP – VQ) is positive. Also VB is less negative r than VA . Thus, VB > VA or (VB – VA) is positive. (b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, (P.E.)A > (P.E.)B and hence sign of potential energy differences is positive. http://video.mit.edu/watch/4-electrostatic-potential-elctric-energy-ev-conservative-field- (c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative. (d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive. EXAMPLE (e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A. 2.3 53 Reprint 2025-26 Physics 2.6 EQUIPOTENTIAL SURFACES An equipotential surface is a surface with a constant value of potential at all points on the surface. For a single charge q, the potential is given by Eq. (2.8): 1 q V = 4 πεo r This shows that V is a constant if r is constant. Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge. Now the electric field lines for a single charge q are radial lines starting from or ending at the charge, depending on whether q is positive or negative. Clearly, the electric field at every point is normal to the equipotential surface passing through that point. This is true in general: for any charge configuration, equipotential surface through a point is normal to the electric field at that point. The proof of this statement is simple. If the field were not normal to the equipotential surface, it would have non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would have to be done. But this is in contradiction to the definition of an equipotential FIGURE 2.9 For a surface: there is no potential difference between any two points on the single charge q surface and no work is required to move a test charge on the surface. (a) equipotential The electric field must, therefore, be normal to the equipotential surface surfaces are at every point. Equipotential surfaces offer an alternative visual picture spherical surfaces in addition to the picture of electric field lines around a charge centred at the configuration. charge, and (b) electric field lines are radial, starting from the charge if q > 0. FIGURE 2.10 Equipotential surfaces for a uniform electric field. For a uniform electric field E, say, along the x-axis, the equipotential surfaces are planes normal to the x-axis, i.e., planes parallel to the y-z plane (Fig. 2.10). Equipotential surfaces for (a) a dipole and (b) two identical positive charges are shown in Fig. 2.11. FIGURE 2.11 Some equipotential surfaces for (a) a dipole, 54 (b) two identical positive charges. Reprint 2025-26 Electrostatic Potential and Capacitance 2.6.1 Relation between field and potential Consider two closely spaced equipotential surfaces A and B (Fig. 2.12) with potential values V and V + dV, where dV is the change in V in the direction of the electric field E. Let P be a point on the surface B. d l is the perpendicular distance of the surface A from P. Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. The work done in this process is |E|dl. This work equals the potential difference VA–VB. Thus, |E|d l = V – (V + dV)= – dV V i.e., |E|= −δ (2.20) δl Since dV is negative, dV = – |dV|. we can rewrite FIGURE 2.12 From the Eq (2.20) as potential to the field. δV δV E = − = + (2.21) δl δl We thus arrive at two important conclusions concerning the relation between electric field and potential: (i) Electric field is in the direction in which the potential decreases steepest. (ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.