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.

3.20 For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. t (sec) P(mm of Hg) 0 35.0 360 54.0 720 63.0 Calculate the rate constant.

2.14 COMBINATION OF CAPACITORS We can combine several capacitors of capacitance C1, C2,…, Cn to obtain a system with some effective capacitance C. The effective capacitance depends on the way the individual capacitors are combined. Two simple possibilities are discussed below. 2.14.1 Capacitors in series Figure 2.26 shows capacitors C1 and C2 FIGURE 2.26 Combination of two combined in series. capacitors in series. The left plate of C1 and the right plate of C2 are connected to two terminals of a battery and have charges Q and –Q , respectively. It then follows that the right plate of C1 has charge –Q and the left plate of C2 has charge Q. If this was not so, the net charge on each capacitor would not be zero. This would result in an electric field in the conductor connecting C1and C2. Charge would flow until the net charge on both C1 and C2 is zero and there is no electric field in the conductor connecting C1 and C2. Thus, in the series combination, charges on the two plates FIGURE 2.27 Combination of n (±Q) are the same on each capacitor. The total capacitors in series. 71 Reprint 2025-26 Physics potential drop V across the combination is the sum of the potential drops V1 and V2 across C1 and C2, respectively. Q Q + (2.55) V = V1 + V2 = C1 C 2 V 1 1 i.e., = + , (2.56) Q C1 C 2 Now we can regard the combination as an effective capacitor with charge Q and potential difference V. The effective capacitance of the combination is Q C = (2.57) V We compare Eq. (2.57) with Eq. (2.56), and obtain 1 1 1 = + (2.58) C C1 C 2 The proof clearly goes through for any number of capacitors arranged in a similar way. Equation (2.55), for n capacitors arranged in series, generalises to Q Q Q V = V1 + V 2 + ... + V n = + + ... + (2.59) C1 C 2 C n Following the same steps as for the case of two capacitors, we get the general formula for effective capacitance of a series combination of n capacitors: 1 1 1 1 1 = + + + ... + (2.60) C C1 C 2 C 3 C n 2.14.2 Capacitors in parallel Figure 2.28 (a) shows two capacitors arranged in parallel. In this case, the same potential difference is applied across both the capacitors. But the plate charges (±Q1) on capacitor 1 and the plate charges (±Q2) on the capacitor 2 are not necessarily the same: Q1 = C1V, Q2 = C2V (2.61) The equivalent capacitor is one with charge Q = Q1 + Q2 (2.62) and potential difference V. Q = CV = C1V + C2V (2.63) The effective capacitance C is, from Eq. (2.63), C = C1 + C2 (2.64) The general formula for effective capacitance C for parallel combination of n capacitors [Fig. 2.28 (b)] follows similarly, Q = Q1 + Q2 + ... + Qn (2.65) FIGURE 2.28 Parallel combination of i.e., CV = C1V + C2V + ... CnV(2.66) (a) two capacitors, (b) n capacitors. which gives C = C1 + C2 + ... Cn (2.67) 72 Reprint 2025-26 Electrostatic Potential and Capacitance Example 2.9 A network of four 10 mF capacitors is connected to a 500 V supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.) FIGURE 2.29 Solution (a) In the given network, C1, C2 and C3 are connected in series. The effective capacitance C¢ of these three capacitors is given by 1 1 1 1 = + + C ′ C1 C 2 C 3 For C1 = C2 = C3 = 10 mF, C¢ = (10/3) mF. The network has C¢ and C4 connected in parallel. Thus, the equivalent capacitance C of the network is  10  C = C¢ + C4 =  3 + 10 mF =13.3mF (b) Clearly, from the figure, the charge on each of the capacitors, C1, C2 and C3 is the same, say Q. Let the charge on C4 be Q¢. Now, since the potential difference across AB is Q/C1, across BC is Q/C2, across CD is Q/C3 , we have Q Q Q + + = 500 V . C1 C 2 C 3 Also, Q¢/C4 = 500 V. This gives for the given value of the capacitances, 10 −3 Q = 500 V × µ F = 1.7 × 10 C and EXAMPLE 3 Q ′ = 500 V × 10 µ F = 5.0 × 10 −3 C 2.9