PHYS-102SPRING--08HW-04

PHYS-102SPRING--08HW-04 - PHYS-102 P20.41 (a) 1 1 1 = + Cs...

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PHYS-102 Homework-04 Solutions Electric Potential and Capacitance P20.41 (a) 1 C s = 1 15.0 + 1 3.00 C s = 2.50 μ F C p = 2.50 + 6.00 = 8.50 F C eq = 1 8.50 F + 1 20.0 F 1 = 5.96 F (b) Q = C Δ V = 5.96 F () 15.0 V = 89.5 C on 20.0 F Δ V = Q C = 89.5 C 20.0 F = 4.47 V 15.0 4.47 = 10.53 V Q = C Δ V = 6.00 F 10.53 V = 63.2 C on 6.00 F 89.5 63.2 = 26.3 C on 15.0 F and 3.00 FIG. P20.41 P20.49 U = 1 2 C Δ V 2 The circuit diagram is shown at the right. (a) C p = C 1 + C 2 = 25.0 F + 5.00 F = 30.0 U = 1 2 30.0 × 10 6 100 2 = 0.150 J (b) C s = 1 C 1 + 1 C 2 1 = 1 25.0 F + 1 5.00 F 1 = 4.17 U = 1 2 C Δ V 2 Δ V = 2 U C = 20 .150 4.17 × 10 6
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This note was uploaded on 09/03/2008 for the course PHYS 102 taught by Professor N/a during the Spring '08 term at Drexel.

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