2.7 POTENTIAL ENERGY OF A SYSTEM OF CHARGES Consider first the simple case of two charges q1and q2 with position vector r1 and r2 relative to some origin. Let us calculate the work done (externally) in building up this configuration. This means that we consider the charges q1 and q2 initially at infinity and determine the work done by an external agency to bring the charges to the given locations. Suppose, first the charge q1 is brought from infinity to the point r1. There is no external field against which work needs to be done, so work done in bringing q1 from infinity to r1 is zero. This charge produces a potential in space given by 1 q1 V1 = 4 πε0 r1P where r1P is the distance of a point P in space from the location of q1. From the definition of potential, work done in bringing charge q2 from infinity to the point r2 is q2 times the potential at r2 due to q1: 1 q1q 2 work done on q2 = 4 πε0 r12 55 Reprint 2025-26 Physics where r12 is the distance between points 1 and 2. Since electrostatic force is conservative, this work gets stored in the form of potential energy of the system. Thus, the potential energy of a system of two charges q1 and q2 is FIGURE 2.13 Potential energy of a 1 q1q 2 U = system of charges q1 and q2 is 4 πε0 r12 (2.22) directly proportional to the product of charges and inversely to the Obviously, if q2 was brought first to its present location and distance between them. q1 brought later, the potential energy U would be the same. More generally, the potential energy expression, Eq. (2.22), is unaltered whatever way the charges are brought to the specified locations, because of path-independence of work for electrostatic force. Equation (2.22) is true for any sign of q1and q2. If q1q2 > 0, potential energy is positive. This is as expected, since for like charges (q1q2 > 0), electrostatic force is repulsive and a positive amount of work is needed to be done against this force to bring the charges from infinity to a finite distance apart. For unlike charges (q1 q2 < 0), the electrostatic force is attractive. In that case, a positive amount of work is needed against this force to take the charges from the given location to infinity. In other words, a negative amount of work is needed for the reverse path (from infinity to the present locations), so the potential energy is negative. Equation (2.22) is easily generalised for a system of any number of point charges. Let us calculate the potential energy of a system of three charges q1, q2 and q3 located at r1, r2, r3, respectively. To bring q1 first from infinity to r1, no work is required. Next we bring q2 from infinity to r2. As before, work done in this step is 1 q1q 2 q 2 V1 ( r2 ) = (2.23) 4 πε0 r12 The charges q1 and q2 produce a potential, which at any point P is given by 1  q1 q 2  V1, 2 = + (2.24) 4 πε0  r1P r2 P  Work done next in bringing q3 from infinity to the point r3 is q3 times V1, 2 at r3 1  q1q 3 q 2 q 3  q 3 V1, 2 ( r3 ) = + (2.25) 4 πε0  r13 r23  The total work done in assembling the charges at the given locations is obtained by adding the work done in different steps [Eq. (2.23) and Eq. (2.25)], 1  q1q 2 q1q 3 q 2 q 3  U = + + (2.26) FIGURE 2.14 Potential energy of a 4 πε0  r12 r13 r23  system of three charges is given by Again, because of the conservative nature of the Eq. (2.26), with the notation given electrostatic force (or equivalently, the path in the figure. independence of work done), the final expression for U, Eq. (2.26), is independent of the manner in which 56 the configuration is assembled. The potential energy Reprint 2025-26 Electrostatic Potential and Capacitance is characteristic of the present state of configuration, and not the way the state is achieved. Example 2.4 Four charges are arranged at the corners of a square ABCD of side d, as shown in Fig. 2.15.(a) Find the work required to put together this arrangement. (b) A charge q0 is brought to the centre E of the square, the four charges being held fixed at its corners. How much extra work is needed to do this? FIGURE 2.15 Solution (a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges –q, +q, and –q are brought to B, C and D, respectively. The total work needed can be calculated in steps: (i) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero. (ii) Work needed to bring –q to B when +q is at A. This is given by (charge at B) × (electrostatic potential at B due to charge +q at A)  q  q 2 = −q × − ε  4 πε0 d = 4 π 0 d (iii) Work needed to bring charge +q to C when +q is at A and –q is at B. This is given by (charge at C) × (potential at C due to charges at A and B)  + q −q  = + q + ε 0 d   4 πε0 d 2 4 π −q 2  1  = 4 πε0 d 1 − 2  (iv) Work needed to bring –q to D when +q at A,–q at B, and +q at C. This is given by (charge at D) × (potential at D due to charges at A, B and C)  + q −q q  = −q + + ε  4 π 0 d 4 πε0 d 2 4 πε0 d  EXAMPLE −q 2  1  = 4 πε0 d  2 − 2  2.4 57 Reprint 2025-26 Physics Add the work done in steps (i), (ii), (iii) and (iv). The total work required is −q 2   1   1   =  4 πε0 d ( 0 ) + (1) + 1 − 2 +  2 − 2    −q 2 = 4 − 2 ( ) 4 πε0 d The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges. (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.) 2.4 the(b) Thefourextrachargesworkarenecessaryat A, B, CtoandbringD ais chargeq0 × (electrostaticq0 to the pointpotentialE whenat E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D. Hence, no work is required to bring any charge to EXAMPLE point E. 2.8 POTENTIAL ENERGY IN AN EXTERNAL FIELD 2.8.1 Potential energy of a single charge In Section 2.7, the source of the electric field was specified – the charges and their locations - and the potential energy of the system of those charges was determined. In this section, we ask a related but a distinct question. What is the potential energy of a charge q in a given field? This question was, in fact, the starting point that led us to the notion of the electrostatic potential (Sections 2.1 and 2.2). But here we address this question again to clarify in what way it is different from the discussion in Section 2.7. The main difference is that we are now concerned with the potential energy of a charge (or charges) in an external field. The external field E is not produced by the given charge(s) whose potential energy we wish to calculate. E is produced by sources external to the given charge(s).The external sources may be known, but often they are unknown or unspecified; what is specified is the electric field E or the electrostatic potential V due to the external sources. We assume that the charge q does not significantly affect the sources producing the external field. This is true if q is very small, or the external sources are held fixed by other unspecified forces. Even if q is finite, its influence on the external sources may still be ignored in the situation when very strong sources far away at infinity produce a finite field E in the region of interest. Note again that we are interested in determining the potential energy of a given charge q (and later, a system of charges) in the external field; we are not interested in the potential energy of the sources producing the external electric field. The external electric field E and the corresponding external potential V may vary from point to point. By definition, V at a point P is the work 58 done in bringing a unit positive charge from infinity to the point P. Reprint 2025-26 Electrostatic Potential and Capacitance (We continue to take potential at infinity to be zero.) Thus, work done in bringing a charge q from infinity to the point P in the external field is qV. This work is stored in the form of potential energy of q. If the point P has position vector r relative to some origin, we can write: Potential energy of q at r in an external field = qV(r) (2.27) where V(r) is the external potential at the point r. Thus, if an electron with charge q = e = 1.6×10–19 C is accelerated by a potential difference of DV = 1 volt, it would gain energy of qDV = 1.6 × 10–19J. This unit of energy is defined as 1 electron volt or 1eV, i.e., 1 eV=1.6 × 10–19J. The units based on eV are most commonly used in atomic, nuclear and particle physics, (1 keV = 103eV = 1.6 × 10–16J, 1 MeV = 106eV = 1.6 × 10–13J, 1 GeV = 109eV = 1.6 × 10–10J and 1 TeV = 1012eV = 1.6 × 10–7J). [This has already been defined on Page 117, XI 2.8.2 Potential energy of a system of two charges in an external field Next, we ask: what is the potential energy of a system of two charges q1 and q2 located at r1and r2, respectively, in an external field? First, we calculate the work done in bringing the charge q1 from infinity to r1. Work done in this step is q1 V(r1), using Eq. (2.27). Next, we consider the work done in bringing q2 to r2. In this step, work is done not only against the external field E but also against the field due to q1. Work done on q2 against the external field = q2 V (r2) Work done on q2 against the field due to q1 q1q 2 = 4 πεo r12 where r12 is the distance between q1 and q2. We have made use of Eqs. (2.27) and (2.22). By the superposition principle for fields, we add up the work done on q2 against the two fields (E and that due to q1): Work done in bringing q2 to r2 q1q 2 = q 2 V ( r2 ) + (2.28) 4 πεo r12 Thus, Potential energy of the system = the total work done in assembling the configuration q1q 2 = q1V ( r1 ) + q 2 V ( r2 ) + (2.29) 4 πε0r12 Example 2.5 (a) Determine the electrostatic potential energy of a system consisting of two charges 7 mC and –2 mC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively. EXAMPLE (b) How much work is required to separate the two charges infinitely 2.5 away from each other? 59 Reprint 2025-26 Physics (c) Suppose that the same system of charges is now placed in an external electric field E = A (1/r 2); A = 9 × 105 NC–1 m2. What would the electrostatic energy of the configuration be? Solution 1 q1q 2 9 7 × ( −2) × 10 −12 (a) U = = 9 × 10 × = –0.7 J. 4 πε0 r 0.18 (b) W = U2 – U1 = 0 – U = 0 – (–0.7) = 0.7 J. (c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find, 7 µ C −µ2 C q1V ( r1 ) + q 2 V ( r2 ) = A + A 0.09m 0.09m 2.5 and the net electrostatic energy is q1q 2 7 µC −µ2 C q1V ( r1 ) + q 2 V ( r2 ) + = A + A − 0.7 J 4 πε0r12 0.09 m 0.09 m EXAMPLE = 70 − 20 − 0.7 = 49.3 J 2.8.3 Potential energy of a dipole in an external field Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform electric field E, as shown in Fig. 2.16. As seen in the last chapter, in a uniform electric field, the dipole experiences no net force; but experiences a torque ttttt given by ttttt ===== p × E (2.30) which will tend to rotate it (unless p is parallel or antiparallel to E). Suppose an external torque text is applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle q0 to angle q1 at an infinitesimal angular speed and without angular acceleration. The amount of work done by the external torque will be given by FIGURE 2.16 Potential energy of a dipole in a uniform external field. = pE ( cosθ0 − cosθ1 ) (2.31) This work is stored as the potential energy of the system. We can then associate potential energy U(q) with an inclination q of the dipole. Similar to other potential energies, there is a freedom in choosing the angle where the potential energy U is taken to be zero. A natural choice is to take q0 = p / 2. (An explanation for it is provided towards the end of discussion.) We can then write, (2.32) 60 Reprint 2025-26 Electrostatic Potential and Capacitance This expression can alternately be understood also from Eq. (2.29). We apply Eq. (2.29) to the present system of two charges +q and –q. The potential energy expression then reads q 2 U ′ (θ) = q [V ( r1 ) − V ( r2 )] − (2.33) 4 πε0 × 2a Here, r1 and r2 denote the position vectors of +q and –q. Now, the potential difference between positions r1 and r2 equals the work done in bringing a unit positive charge against field from r2 to r1. The displacement parallel to the force is 2a cosq. Thus, [V(r1)–V (r2)] = –E × 2a cosq . We thus obtain, q 2 q 2 U ′ (θ) = − pE cosθ− = − p.E − (2.34) 4 πε0 × 2a 4 πε0 × 2a We note that U¢(q) differs from U(q ) by a quantity which is just a constant for a given dipole. Since a constant is insignificant for potential energy, we can drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32). We can now understand why we took q0=p/2. In this case, the work done against the external field E in bringing +q and – q are equal and opposite and cancel out, i.e., q [V (r1) – V (r2)]=0. Example 2.6 A molecule of a substance has a permanent electric dipole moment of magnitude 10–29 C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 106 V m–1. The direction of the field is suddenly changed by an angle of 60º. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample. Solution Here, dipole moment of each molecules = 10–29 C m As 1 mole of the substance contains 6 × 1023 molecules, total dipole moment of all the molecules, p = 6 × 1023 × 10–29 C m = 6 × 10–6 C m Initial potential energy, Ui = –pE cos q = –6×10–6×106 cos 0° = –6 J Final potential energy (when q = 60°), Uf = –6 × 10–6 × 106 cos 60° = –3 J Change in potential energy = –3 J – (–6J) = 3 J EXAMPLE So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles. 2.6

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.

2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.