# 2_8 - PHY 5346 HW Set 4 Solutions Kimel 1 2.8 The system is...

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Unformatted text preview: PHY 5346 HW Set 4 Solutions Kimel 1. 2.8 The system is pictured below a) Using the known potential for a line charge, the two line charges above give the potential 1 ln r V, a constant. Let us define V 4 0 V r r 2 0 Then the above equation can be written r 2 e V or r 2 r 2 e V r 2 Writing r 2 R , the above can be written r 2 z r e R V 1 R e V R2e e 1 V V 1 2 Re 2 e V V The equation is that of a circle whose center is at z , and whose radius is a 1 b) The geometry of the system is shown in the fugure. Note that d R d1 d2 with d1 e and a e Forming Re 2 Va Va R Va , d2 1 R e Vb 1 , b 1 e Re Vb Vb 2 1 2 d or 2 a 2 b 2 R e R Va 1 R e Vb Re 2 1 Va 2 Re 1 e Vb Vb 2 2 e R2 e Va Vb Va 1 1 Vb d2 Thus we can write d2 or Va a2 b2 2ab e a2 b2 e Va 1 e 1 Va Vb 1 Vb 2 e Va Vb 2 2e 2 e Va e 2 d2 Va Vb 2 cosh Va Vb 2 Vb 1 2 cosh 0 1 a2 b2 2ab Q/L 2 0 Capacitance/unit length C 1 d2 a2 b2 Va Vb L Va Vb cosh 2ab c) Suppose a 2 d 2 , and b 2 d 2 , and a ab , then 2 0 2 C d2 1 1 d2 a2 b2 L cosh cosh 1 2a 2 cosh d2 1 or ln or C L Let us defind 2 a 2 b 2 /d 2 , then 2 0 C d2 1 2 L ln a2 ln 2 0 d 2 1 a 2 b 2 /d 2 a2 1 0 a 2 b 2 /d 2 2a 2 d2 1 e a 2 b 2 /d 2 2a 2 2 0 L C 2 0 L C a 2 b 2 /d 2 2a 2 d2 1 2 negligible terms if 2 0 L 1 C 2 0 L C a 2 b 2 /d 2 a2 2 ln 0 d2 a2 2 ln 2 0 d2 a2 2 O 4 The first term of this result agree with problem 1.7, and the second term gives the appropriate correction asked for. d) In this case, we must take the opposite sign for d 2 a 2 b 2 , since a 2 b 2 d 2 . Thus 2 0 C 1 a 2 b 2 d 2 L cosh 2a 2 If we use the identiy, ln x x 2 1 cosh 1 x , G.&R., p. 50., then for d 0 C L ln in agreement with problem 1.6. a 2 b 2 2ab 2 0 2 b2 a2ab 2 0 ln a b ...
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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.

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