chapt25 - Chapter 25: Capacitance Problem 2 The capacitor...

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Chapter 25: Capacitance Problem 2 The capacitor in Fig. 25-26 has a capacitance of 25 μ F and is initially uncharged. The battery provides a potential difference of 120 V. After switch S is closed, how much charge will pass through it? Answer 3.0 mC. S C + _ Figure 25-26 Solution After the switch has been closed a long time the potential difference across the capacitor will be the same as the potential difference across the battery. Therefore the capacitor will now have a charge of q = CV = (25 μ F)*(120 V) = 3.0 mC. This charge will have passed through the battery and switch in the process of charging the capacitor. Problem 5 A parallel-plate capacitor has circular plates of 8.20-cm radius and 1.30-mm separation. (a) Calculate the capacitance. (b) What charge will appear on the plates if a potential difference of 120 V is applied? Answer (a) 141 pF. (b) 16.9 nC. (a) The capacitance of a parallel plate capacitor of area A and plate separation d is C = ε o A d = 8.85 × 10 12 F m π 8.20 cm ( ) 2 1.30 mm = 141 pF . (b) The charge on the capacitor will be q = CV = (141 pF)*(120 V) = 16.9 nC. Problem 6 The plates of a spherical capacitor have radii 38.0 and 40.0 mm. (a) Calculate the capacitance. (b) What must be the plate area of a parallel-plate capacitor with the same plate separation and capacitance? Answer (a) 84.5 pF. (b) 191 cm 2 .
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25-2 Chapter 25: Capacitance Solution (a) The capacitance of a spherical capacitor with plates of inner radius a and outer radius b is given by C = 4 πε o ab b a = 4 o (38 mm )*(40 mm ) 40 mm 38 mm = 84.5 pF (b) The capacitance of a parallel plate capacitor is C = ε o A d Therefore the area is A = Cd o == 84.5 pF 2 mm 8.85 × 10 12 F m = 191 cm 2 . Problem 8 In Fig. 25-28 find the equivalent capacitance of the combination. Assume that C 1 = 10.0 μ F, C 2 = 5.00 μ F, and C 3 = 4.00 μ F. Answer 7.33 μ F. V C 1 C 2 C 3 Figure 25-28 Capacitors C 1 and C 2 are in series and their equivalent capacitance, C 1,2 , is C 1,2 = C 1 C 2 C 1 + C 2 = 10 μ F 5 F 10 F + 5 F = 3.33 F . This equivalent capacitance is in parallel with C 3 therefore the total equivalent capacitance, C tot C tot = C 1,2 + C 3 = 3.33 F + 4 F = 7.33 F Problem 10 How many 1.00- μ F capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the capacitors? Answer 9091. The total capacitance of parallel combination must be
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Chapter 25: Capacitance 25-3 C tot = q V = 1.00 C 110 V = 9091 μ F For parallel combinations of capacitors C tot = Σ C i . In this case we are combining n identical 1- μ F capacitors. Therefore Σ C i = nC and n, the number of capacitors, must be 9091. Problem 11 Each of the uncharged capacitors in Figure 25-30 has a capacitance of 25.0 μ F. A potential difference of 4200 V is established when the switch is closed. How many coulombs of charge then pass through the meter A? A C C C V Figure 25-30 Answer 315 mC. Solution The total charge that will pass through meter A is the sum of the charges on the three capacitors after they have a potential difference across each of 4200 V. The charge on one of the capacitors is q = CV = (25 μ F)*(4200 V) = 105 mC.
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chapt25 - Chapter 25: Capacitance Problem 2 The capacitor...

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