PHYS-102hw4soln

# PHYS-102hw4soln - P20.69 From Example 20.5 the potential at...

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PHYS-102 Homework 4 Solutions Spring -08 ____________________________________________________________ Chapter 20 Electric Potential and Capacitance 41, 49, 69 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 ! V = 5.96 F ( ) 15.0 V ( ) = 89.5 C on 20.0 F ! V = Q C = 89.5 C 20.0 F = 4.47 V 15.0 " 4.47 = 10.53 V Q = C ! 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 F FIG. P20.41

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P20.49 U = 1 2 C ! V ( ) 2 The circuit diagram is shown at the right. (a) C p = C 1 + C 2 = 25.0 μ F + 5.00 F = 30.0 F 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 F U = 1 2 C ! V ( ) 2 ! V = 2 U C = 2 0.150 ( ) 4.17 " 10 # 6 = 268 V
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Unformatted text preview: P20.69 From Example 20.5, the potential at the center of the ring is V i = k e Q R and the potential at an infinite distance from the ring is V f = . Thus, the initial and final potential energies of the point charge-ring system are: U i = QV i = k e Q 2 R and U f = QV f = . From conservation of energy, K f + U f = K i + U i or 1 2 Mv f 2 + = + k e Q 2 R giving v f = 2 k e Q 2 MR ....
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PHYS-102hw4soln - P20.69 From Example 20.5 the potential at...

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