100 f300 f c c v q q c c μ μ μ μ μ b capacitors

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1.00 F+3.00 F C C V q q C C μ μ μ μ μ = = = = + (b) Capacitors 2 and 4 are also in series: ( ( ( 2 4 2 4 2 4 2.00 F 4.00 F 12.0V 16.0 C. 2.00 F 4.00 F C C V q q C C μ μ μ μ μ = = = = + + (c) 3 1 9.00 C. q q μ = = (d) 4 2 16.0 C. q q μ = = (e) With switch 2 also closed, the potential difference V 1 across C 1 must equal the potential difference across C 2 and is ( ( 3 4 1 1 2 3 4 3.00 F 4.00 F 12.0V 8.40V. 1.00 F 2.00 F 3.00 F 4.00 F C C V V C C C C μ μ μ μ μ μ + + = = = + + + + + + Thus, q 1 = C 1 V 1 = (1.00 μ F)(8.40 V) = 8.40 μ C. (f) Similarly, q 2 = C 2 V 1 = (2.00 μ F)(8.40 V) = 16.8 μ C. (g) q 3 = C 3 ( V – V 1 ) = (3.00 μ F)(12.0 V – 8.40 V) = 10.8 μ C. (h) q 4 = C 4 ( V – V 1 ) = (4.00 μ F)(12.0 V – 8.40 V) = 14.4 μ C. 28. Initially the capacitors C 1 , C 2 , and C 3 form a combination equivalent to a single capacitor which we denote C 123 . This obeys the equation 1 2 3 123 1 2 3 1 2 3 1 1 1 ( ) C C C C C C C C C C + = + = + + . Hence, using q = C 123 V and the fact that q = q 1 = C 1 V 1 , we arrive at 123 2 3 1 1 1 1 1 1 2 3 C C C q q V V V C C C C C C = = = = + + . (a) As C 3 this expression becomes V 1 = V . Since the problem states that V 1 approaches 10 volts in this limit, so we conclude V = 10 V.
1029 (b) and (c) At C 3 = 0, the graph indicates V 1 = 2.0 V. The above expression consequently implies C 1 = 4 C 2 . Next we note that the graph shows that, at C 3 = 6.0 μ F, the voltage across C 1 is exactly half of the battery voltage. Thus, 1 2 = C 2 + 6.0 μ F C 1 + C 2 + 6.0 μ F = C 2 + 6.0 μ F 4 C 2 + C 2 + 6.0 μ F which leads to C 2 = 2.0 μ F. We conclude, too, that C 1 = 8.0 μ F. 29. The total energy is the sum of the energies stored in the individual capacitors. Since they are connected in parallel, the potential difference V across the capacitors is the same and the total energy is ( 29 ( 29 ( 29 2 2 6 6 1 2 1 1 2.0 10 F 4.0 10 F 300V 0.27 J. 2 2 U C C V - - = + = × + × = 30. (a) The capacitance is ( ( 12 2 2 4 2 11 0 3 8.85 10 C /N m 40 10 m 3.5 10 F 35pF. 1.0 10 m A C d ε - - - - × × = = = × = × (b) q = CV = (35 pF)(600 V) = 2.1 × 10 –8 C = 21 nC. (c) U CV = = = × - 1 2 2 1 2 2 6 35 21 63 10 pF nC J = 6.3 J. b gb g . μ (d) E = V / d = 600 V/1.0 × 10 –3 m = 6.0 × 10 5 V/m. (e) The energy density (energy per unit volume) is ( 29( 29 6 3 4 2 3 6.3 10 J 1.6 J m . 40 10 m 1.0 10 m U u Ad - - - × = = = × × 31. The energy stored by a capacitor is given by U CV = 1 2 2 , where V is the potential difference across its plates. We convert the given value of the energy to Joules. Since 1 J 1 W s, = we multiply by (10 3 W/kW)(3600 s/h) to obtain 7 10 kW h 3.6 10 J = × . Thus, C U V = = × = 2 2 36 10 1000 72 2 7 2 . J V F. c h b g 32. Let ς = 1.00 m 3 . Using Eq. 25-25, the energy stored is
CHAPTER 25 1030 ( 29 ( 29 2 2 2 12 3 8 0 2 1 1 C 8.85 10 150V m 1.00m 9.96 10 J. 2 2 N m U u E ε - - = = = × = × V V 33. The energy per unit volume is u E e r e r = = F H G I K J = 1 2 1 2 4 32 0 2 0 2 2 2 0 4 ε ε ε ε π π 0 2 . (a) At 3 1.00 10 m r - = × , with 19 1.60 10 C e - = × and 12 2 2 0 8.85 10 C /N m ε - = × , we have 18 3 9.16 10 J/m u - = × . (b) Similarly, at 6 1.00 10 m r - = × , 6 3 9.16 10 J/m u - = × . (c) At 9 1.00 10 m r - = × , 6 3 9.16 10 J/m u = × . (d) At 12 1.00 10 m r - = × , 18 3 9.16 10 J/m u = × . (e) From the expression above u r –4 . Thus, for r 0, the energy density u .

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