Unformatted text preview: PHY 5346 HW Set 4 Solutions Kimel 4. 2.11 The system is pictured in the following figure: a) The potential for a line charge is (see problem 2.3) ln rr r 2 0 Thus for this system r 2 ln
0 r R r 2 ln
0 r R r To determine and R , we need two conditions: I) As r , we want b r 0, so . or ln b R R b or b R R b This is an equation for R with the solution bR bR ln b R bR II) b r 2 R b R The same condition we found for a sphere. b) r 4 as r r Using 4 ln
0 ln
0 r2 b4 R2 2r b cos R 2rR cos 2 b2 rR 2 r2 R2 4 1 2b 2 cos /rR 1 2R cos /r ln 1 x x r 2 4 2b
0 ln 1
0 R 2 cos 1 x2 1 x3 O x4 3 2 1 b 2 R 2 cos rR r2 b4 R2 c) 0 r rb r ln 2r b cos R 2rR cos rb 2 r2 R2
2 1 y y 1 2y cos 2 where y R/b. Plotting / 2b 1 y2 y 2 1 2y cos , for y 2, 4, gives gy g 2 ,g 4 3 5 4 cos 1 y2 y 2 1 2y cos , 15 17 8 cos 0.5 0 0.5 1 1.5 2 2.5 3 1 1.5 2 2.5 3 d) If the line charges are a distance d apart, then the electric field at from is, using Gauss's law E 2 0 d The force on is LE, ie, 2 L , and the force is attractive. F L 2 0 d 2 0 d ...
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This note was uploaded on 01/10/2012 for the course PHYSIS 506 taught by Professor Smith during the Spring '08 term at University of Arkansas Community College at Morrilton.
 Spring '08
 smith

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