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Course: PHYS 1200, Spring 2008
School: RPI
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PHYSICS PHYS-1200 II HOMEWORK SOLUTIONS CLASS 4. 25.03, 25.23, 25.67; Q26.02, Q26.05 25.03. a) C SPRING 2006 0 A d 0 r2 d (8.85 10 -10 12 (0.0820m)2 F/m) 1.30 10 3 m C 1.44 10 10 F 144 pF Q = 1.73 10 C = 17.3 nC -8 b) Q = CV = (1.44 10 F)(120 V) 25.23. When only switch S1 is closed, The series combination of C1 and C3 is connected across the battery, as is the series combination of C2 and C4....

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PHYSICS PHYS-1200 II HOMEWORK SOLUTIONS CLASS 4. 25.03, 25.23, 25.67; Q26.02, Q26.05 25.03. a) C SPRING 2006 0 A d 0 r2 d (8.85 10 -10 12 (0.0820m)2 F/m) 1.30 10 3 m C 1.44 10 10 F 144 pF Q = 1.73 10 C = 17.3 nC -8 b) Q = CV = (1.44 10 F)(120 V) 25.23. When only switch S1 is closed, The series combination of C1 and C3 is connected across the battery, as is the series combination of C2 and C4. Consider each series combination separately. For C1 and C3, the charge on each capacitor is the same, and equal to the charge on the series combination. Calculate the capacitance of the series combination, and then calculate the charge on the combination. C13 C1C3 , so C1 C3 q13 C13V C1C3 V C1 C3 (1.00 F)(3.00 F) (12.0 V) 1.00 F 3.00 F 9.00 C Do the same for C2 and C4. C24 a) b) c) d) C2 C4 , so C2 C4 q24 C24V C2C4 V C2 C4 (2.00 F)(4.00 F) (12.0 V) 16.0 C 2.00 F 4.00 F q1 = q13 = 9.00 q2 = q24 = 16.0 q3 = q13 = 9.00 q4 = q24 = 16.0 C C C C When both S1 and S2 are closed, the parallel combination of C1 and C2 is in series with the parallel combination of C3 and C4. Since C1 and C2 have the same voltage across them, that voltage must be found, then the charge on each capacitor can be calculated. The same must be done for C1 and C2. Call the voltage across C1 and C2, V12, and that across C3 and C4, V34. Also the charge on the combination of C1 and C2, must equal that on the combination of C3 and C4, since they are in series. Then, V12 + V34 = V, and C12V12 = C34V34 where C12 = C1 + C2, and C34 = C3 + C4. The two equations involving the voltages can solved be for V12 and V34. Multiply the upper equation by C34, and we get, C34V12 + C34V34 = C34V C12V12 C34V34 = 0 Adding these two gives, (C12 + C34) V12 = C34V so C34V (C3 C4 )V (3.00 F 4.00 F)12.0 V V12 C12 C34 C1 C2 C3 C4 1.00 F 2.00 F 3.00 F 4.00 F V34 = V V12 = 12.0 V 8.40 V = 3.6 V. e) f) g) h) q1 = C1V12 = (1.00 q2 = C2V12 = (2.00 q3 = C3V34 = (3.00 q4 = C4V34 = (4.00 F)(8.40 V) F)(8.40 V) F)(3.6 V) F)(3.6 V) 8.40 V q1 = 8.40 q2 = 16.8 q3 = 10.8 q4 = 14.4 C C C C 25.67. The maximum capacitance will occur when there is a maximum area of overlap between the two sets of plates. That maximum area is A. There are n plates, so there are n 1 gaps with a plate on each side. That corresponds to n 1 capacitors connected in parallel, each with an area A, and a separation of d. The capacitance of one on the capacitors is A . Then for n 1 of them in parallel, the capacitance becomes: C1 0 d (n 1) 0 A C (n 1)C1 d Q26.02 Consider the positive direction to be to the right. Then for each case: a) i = ( 7 C/s) = 7 C/s b) i = (3 C/s) + ( 4 C/s) = 7 C/s c) i = (2 C/s) + ( 5 C/s) = 7 C/s d) i = (6 C/s) ( 1 C/s) = 5 C/s Then, a = b = c > d Q26.05. a) E = V/L, so for (1) E = V/L, for (2) E = 2V/2L = V/L, for (3) E = 2V/3L = (2/3)V/L. Therefore they must be ranked, (1) and (2) tied, and (3) lowest. b) J = E/ . Since is the same for all, they are ranked according to E. They must be ranked, (1) and (2) tied, and (3) lowest. c) vd = J/(ne). Since (ne) is the same for all, they are ranked according to J. They must be ranked, (1) and (2) tied, and (3) lowest.
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