PHYS-102SPRING--08HW-03

PHYS-102SPRING--08HW-03 - PHYS-102 P20.1 (a) Homework-03...

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PHYS-102 Homework-03 Solutions 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 ms (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 v e 2 +− 1.60 × 10 19 C 120 J C v e = 6.49 × 10 6 P20.12 (a) V = k e q 1 r 1 + k e q 2 r 2 = 2 k e q r V = 2 8.99 × 10 9 N m 2 C 2 2.00 × 10 6 C ( ) 1.00 m 2 + 0.500 m 2
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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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