13.5 The Q value of a nuclear reaction A + b ® C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 11 H+13 H →12 H+12 H (ii) 126 C+126 C →1020 Ne+ 24 He Atomic masses are given to be m ( 12 H ) = 2.014102 u m ( 13 H) = 3.016049 u m ( 126 C ) = 12.000000 u m ( 1020 Ne ) = 19.992439 u

13.2 Obtain the binding energy of the nuclei 5626Fe and 20983 Bi in units of MeV from the following data: m ( 5626Fe ) = 55.934939 u m ( 20983 Bi ) = 208.980388 u

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