physics_6c_sp09_final_exam_solns

physics_6c_sp09_final_exam_solns - Physics 6C Spring 2009...

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Physics 6C Spring 2009 Final Exam Solutions 1. (35 points) A very long cylinder of radius R has a uniform charge density ρ . Find the electric field a) for r > R b) for r < R. R r 1 r 2 L a) By symmetry, field will be radial, perpendicular to axis of cylinder. Using larger Gauss’s law cylinder, E 2 π r 2 L = ρπ R 2 L/ ε 0 E = ρ R 2 /(2 ε 0 r 2 ) b) Again by symmetry, field will be radial, perpendicular to axis of cylinder. Using smaller Gauss’s law cylinder, E 2 π r 1 L = ρπ r 1 2 L/ ε 0 E = ρ r 1 /(2 ε 0 )
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2. (30 points) 3.00 Ω 6.00 Ω b a) Find the equivalent resistance between points a and b . b) If the potential drop between a and b is 10.0 volts, find the current in each resistor. c) If the 2.00 Ω resistor is replaced by a wire, what power is dissipated in the 3.00 Ω resistor? a) 3.00 Ω + 1/(1/6.00 Ω + 1/2.00 Ω ) = 4.5 Ω b) Total current I = 10.0V/4.5 Ω = 2.22A. That is the current in 3.00 Ω resistor. The voltage drop across that resistor is 2.22A x 3.00 Ω = 6.66V, leaving 3.33 V across the other two resistors. So the current in the 6.00 Ω resistor is 3.33V/6.00
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This note was uploaded on 03/08/2011 for the course PHYS 6c taught by Professor Smith,d during the Spring '08 term at UCSC.

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physics_6c_sp09_final_exam_solns - Physics 6C Spring 2009...

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