test1 07solutions

# test1 07solutions - P114 CLASS TEST#1 0800-0930 hours 20...

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P114 CLASS TEST #1 Prof. Cline 0800-0930 hours, 20 February 2007 PRINT NAME UR ID# CIRCLE THE NAME OF YOUR WORKSHOP LEADER: Andrew Mosher Amrita Urdhwareshe Xu Wang Andreas Gennis Michael Hartglass Dylan Premdergast Jonathan Widawsky 1. ________ 2. ________ 3. ________ TOTAL _________ 4. ________ 1 . DONOTSITINASEATTHATISADJACENTTOANOTHERSTUDENT . 2. THIS IS A CLOSED BOOK EXAM 3. Print your name on each sheet 4. Do all problems - Problems 1-4 count for 25 points each. 5. Answer on the sheet containing the problem - use the back if necessary. 6. Making a diagram when appropriate usually aids in solving the problem. 7. Circle your answers and give units when appropriate. 8. Necessary constants and formulae are given on the last page of the exam. 9. Calculators are allowed. 10. Show all steps to get full credit 1

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PRINT NAME 1a:(5pts) Explain what basic experimental facts of electrostatics underlie Gauss’s Law of electrostatics I Closed surface −→ E · d S = 1 ε 0 Z enclosed volume ρdτ = ( Enclosed charge ) ε o 1b:(5pts) Explain what basic experimental facts underlie the circulation relation of electrostatics. I Closed loop C E · d l =0 1c(5pts): Consider a sphere of charge of radius R having uniform volume charge density ρC/m 3 Use the above two relations to derive the electric f eld for r>R 1d(5pts): Use the above two relations to derive the electric f eld for r<R 1e(5pts) Use the electric f eld to calculate the electric potential for r = R assuming V at r = ? Show all steps to get full credit . SOLUTIONS:1a) Coulomb’s Law that E 1 r 2 E is radial Superposition 1b) The Electric f eld is conservative because the electric f eld for a point charge is radial 1c) Use Gauss’s law for a concentric Gaussian sphere with . By symmetry the electric f eld must be radial and uniform on the surface of the sphere. That is along a concentric circle I Closed loop C E · d l since E and d l are perpendicular and thus the dot product is zero. Using Gauss’s law we have that I Closed surface E · d S = 1 ε 0 Z enclosed volume ρdτ E 4 πr 2 = 1 ε 0 4 3 πR 3 ρ ˆ r E = 1 ε 0 1 4 2 4 3 3 ρ ˆ r E = 1 ε 0 Q 4 2 ˆ r 1d) For we have that I Closed surface E · d S = 1 ε 0 Z enclosed volume ρdτ E 4 2 = 1 ε 0 4 3 3 ρ ˆ r E = 1 ε 0 1 4 π 4 3 πrρ ˆ r E = 1 ε 0 Q 4 3 ¯ r 2
1e) For r R we have that the electric potential is V = Z r −→ E ·

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test1 07solutions - P114 CLASS TEST#1 0800-0930 hours 20...

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