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ISM_CH25

# ISM_CH25 - Chapter 25 1 Charge flows until the potential...

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Chapter 25 1. Charge flows until the potential difference across the capacitor is the same as the potential difference across the battery. The charge on the capacitor is then q = CV , and this is the same as the total charge that has passed through the battery. Thus, q = (25 × 10 –6 F)(120 V) = 3.0 × 10 –3 C. 2. (a) The capacitance of the system is C q V = = = 70 20 35 pC V pF. . (b) The capacitance is independent of q ; it is still 3.5 pF. (c) The potential difference becomes V q C = = = 200 35 57 pC pF V. . 3. (a) The capacitance of a parallel-plate capacitor is given by C = ε 0 A / d , where A is the area of each plate and d is the plate separation. Since the plates are circular, the plate area is A = π R 2 , where R is the radius of a plate. Thus, ( 29 ( 29 2 12 2 2 10 0 3 8.85 10 F m 8.2 10 m 1.44 10 F 144pF. 1.3 10 m R C d π ε π - - - - × × = = = × = × (b) The charge on the positive plate is given by q = CV , where V is the potential difference across the plates. Thus, q = (1.44 × 10 –10 F)(120 V) = 1.73 × 10 –8 C = 17.3 nC. 4. We use C = A ε 0 / d . (a) Thus, d A C = = × = × - - ε 0 12 12 100 885 10 100 885 10 2 . . . . m F m. 2 C N m 2 c hd i 1039

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CHAPTER 25 (b) Since d is much less than the size of an atom ( 10 –10 m), this capacitor cannot be constructed. 5. Assuming conservation of volume, we find the radius of the combined spheres, then use C = 4 π ε 0 R to find the capacitance. When the drops combine, the volume is doubled. It is then V = 2(4 π /3) R 3 . The new radius R' is given by ( 29 3 3 4 4 2 3 3 R R = π π ′ = R R 2 1 3 . The new capacitance is 1 3 0 0 0 4 4 2 5.04 . C R R R ε ε ε = = = π π π With R = 2.00 mm, we obtain ( 29 ( 29 12 3 13 5.04 8.85 10 F m 2.00 10 m 2.80 10 F C π - - - = × × = × . 6. (a) We use Eq. 25-17: C ab b a = - = × - = 4 40 0 38 0 8 99 10 40 0 38 0 84 5 0 9 π ε . . . . . . mm mm mm mm pF. N m C 2 2 b gb g d ib g (b) Let the area required be A . Then C = ε 0 A /( b – a ), or ( 29 ( 29 ( 29 ( 29 2 2 2 12 C 0 N m 84.5pF 40.0mm 38.0mm 191cm . 8.85 10 C b a A ε - - - = = = × 7. The equivalent capacitance is given by C eq = q / V , where q is the total charge on all the capacitors and V is the potential difference across any one of them. For N identical capacitors in parallel, C eq = NC , where C is the capacitance of one of them. Thus, / NC q V = and ( 29 ( 29 3 6 1 00C 9 09 10 110V 1 00 10 F q . N . . VC . - = = = × × 8. The equivalent capacitance is C C C C C C eq F F F F F F. = + + = + + = 3 1 2 1 2 4 00 10 0 500 10 0 500 7 33 . . . . . . μ μ μ μ μ μ b gb g 9. The equivalent capacitance is 16
( 29 ( 29 ( 29 1 2 3 eq 1 2 3 10.0 F 5.00 F 4.00 F 3.16 F. 10.0 F 5.00 F 4.00 F C C C C C C C μ μ μ μ μ μ μ + + = = = + + + + 10. The charge that passes through meter A is q C V CV = = = = eq F V C. 3 3 250 4200 0 315 . . μ b gb g 11. (a) and (b) The original potential difference V 1 across C 1 is ( 29 ( 29 eq 1 1 2 3.16 F 100.0V 21.1V. 10.0 F 5.00 F C V V C C μ μ μ = = = + + Thus V 1 = 100.0 V – 21.1 V = 78.9 V and q 1 = C 1 V 1 = (10.0 μ F)(78.9 V) = 7.89 × 10 –4 C. 12. (a) The potential difference across C 1 is V 1 = 10.0 V. Thus, q 1 = C 1 V 1 = (10.0 μ F)(10.0 V) = 1.00 × 10 –4 C. (b) Let C = 10.0 μ F. We first consider the three-capacitor combination consisting of C 2 and its two closest neighbors, each of capacitance C . The equivalent capacitance of this combination is 2 eq 2 1 50 C C C C . C.

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ISM_CH25 - Chapter 25 1 Charge flows until the potential...

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