Chapter25 - 1. (a) The capacitance of the system is C= q 70...

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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. .
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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 = (25 × 10 –6 F)(120 V) = 3.0 × 10 –3 C.
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3. For a given potential difference V , the charge on the surface of the plate is () qN e n A d 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 qC V = (Eq. 25-1). Combining the two expressions leads to Cd ne AV = . With 14 // 5 . 0 1 0 m / V ss dV d V == × and 28 3 8.49 10 / m n (see, for example, Sample Problem 25-1), 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 −− × × = × .
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() ( ) 21 2 2 2 12 0 1.00m 8.85 10 C /N m 8.85 10 m. 1.00F A d C ε ×⋅ == (b) Since d is much less than the size of an atom ( 10 –10 m), this capacitor cannot be constructed. 4. We use C = A 0 / d . (a) The distance between the plates is
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5. (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, () ( ) 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 επ −− ×× == = × = × (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.
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6. (a) We use Eq. 25-17: C ab ba = = ×− = 4 40 0 38 0 899 10 400 380 84 5 0 9 π ε .. . . mm pF. Nm C 2 2 bg di b g (b) Let the area required be A . Then C = 0 A /( b – a ), or () ( ) 2 2 2 12 C 0 84.5pF 40.0mm 38.0mm 191cm . 8.85 10 Cb a A −− == = ×
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7. 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 () 3 3 44 2 33 RR = ππ ′ = 2 13 . The new capacitance is 00 0 2 5 . 0 4 . CR R R εε ′′ == = π 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 π −− × = × .
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8. The equivalent capacitance is CC eq F FF F. =+ + + = 3 12 400 10 0 500 10 0 733 . .. . µ µµ bg
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9. The equivalent capacitance is () ( )( ) 12 3 eq 123 10.0 F 5.00 F 4.00 F 3.16 F. 10.0 F 5.00 F 4.00 F CCC C µµµ µ ++ == = + +
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10. 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 100C 909 10 110V 1 00 10 F q. N. . VC . == = × ×
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11. The charge that passes through meter A is qCV C V === = eq FV C . 3 3 250 4200 0 315 .. µ b gb g
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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.
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter25 - 1. (a) The capacitance of the system is C= q 70...

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