# ch25 - Chapter 25 1(a The capacitance of the system is C q...

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1011 Chapter 25 1. (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 57 pC pF V. . 2. 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 = (30 10 –6 F)(120 V) = 3.6 10 –3 C. 3. (a) The capacitance of a parallel-plate capacitor is given by C = 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 = R 2 , where R is the radius of a plate. Thus,     2 12 2 2 10 0 3 8.85 10 F m 8.2 10 m 1.44 10 F 144 pF. 1.3 10 m R C d   (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. (a) We use Eq. 25-17:       2 2 0 9 N m C 40.0mm 37.0mm 4 82.3 pF. 8.99 10 40.0mm 37.0mm ab C b a  (b) Let the area required be A . Then C = 0 A /( b – a ), or

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CHAPTER 25 1012        2 12 2 2 0 82.3pF 40.0mm 37.0mm 279cm . 8.85 10 C /N m C b a A 5. Assuming conservation of volume, we find the radius of the combined spheres, then use C = 4  0 R to find the capacitance. When the drops combine, the volume is doubled. It is then V = 2(4 /3) R 3 . The new radius R' is given by   3 3 4 4 2 3 3 R R p p   R R 2 1 3 . The new capacitance is 1 3 0 0 0 4 4 2 5.04 . C R R R p p p With R = 2.00 mm, we obtain    12 3 13 5.04 8.85 10 F m 2.00 10 m 2.80 10 F C . 6. We use C = A 0 / d . (a) The distance between the plates is    2 12 2 2 12 0 1.00m 8.85 10 C /N m 8.85 10 m. 1.00F A d C (b) Since d is much less than the size of an atom ( 10 –10 m), this capacitor cannot be constructed. 7. For a given potential difference V , the charge on the surface of the plate is ( ) q Ne nAd e where d is the depth from which the electrons come in the plate, and n is the density of conduction electrons. The charge collected on the plate is related to the capacitance and the potential difference by q CV (Eq. 25-1). Combining the two expressions leads to C d ne A V . With 14 / / 5.0 10 m/V s s d V d V and 28 3 8.49 10 / m n (see, for example, Sample Problem — “Charging the plates in a parallel-plate capacitor”), we obtain 28 3 19 4 2 (8.49 10 / m )(1.6 10 C)(5.0 10 14 m/V) 6.79 10 F/m C A .
1013 8. 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     3 6 1 00C 9 09 10 110V 1 00 10 F q .

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ch25 - Chapter 25 1(a The capacitance of the system is C q...

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