PHYS-102hw3soln

# PHYS-102hw3soln - v e 2" 1.60 10" 19 C 120 J C v e...

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PHYS-102 Homework 3 Solutions Spring -08 ____________________________________________________________ Chapter 20 Electric Potential and Capacitance 1,11,27 P20.1 (a) Energy of the proton-field system is conserved as the proton moves from high to low potential, which can be defined for this problem as moving from 120 V down to 0 V. K i + U i + ! E mech = K f + U f 0 + qV + 0 = 1 2 mv p 2 + 0 1.60 ! 10 " 19 C ( ) 120 V ( ) 1 J 1 V # C \$ % ( ) = 1 2 1.67 ! 10 " 27 kg ( ) v p 2 v p = 1.52 ! 10 5 m s (b) The electron will gain speed in moving the other way, from V i = 0 to V f = 120 V : K i + U i + ! E mech = K f + U f 0 + 0 + 0 = 1 2 mv e 2 + qV 0 = 1 2 9.11 ! 10 " 31 kg ( )

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Unformatted text preview: v e 2 + " 1.60 ! 10 " 19 C ( ) 120 J C ( ) v e = 6.49 ! 10 6 m s P20.11 V = k q i r i i ! V = 8.99 " 10 9 ( ) 7.00 " 10 # 6 ( ) # 1 0.010 0 # 1 0.010 0 + 1 0.038 7 \$ % & ’ ( ) V = # 1.10 " 10 7 V = # 11.0 MV FIG. P20.11 P20.27 V = dV ! = 1 4 " # dq r ! All bits of charge are at the same distance from O . So V = 1 4 ! " Q R # \$ % & ’ ( = 8.99 ) 10 9 N * m 2 C 2 ( ) + 7.50 ) 10 + 6 C 0.140 m # \$ % & ’ ( = + 1.51 MV ....
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## This note was uploaded on 06/02/2008 for the course PHYS 102 taught by Professor N/a during the Spring '08 term at Drexel.

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PHYS-102hw3soln - v e 2" 1.60 10" 19 C 120 J C v e...

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