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Unformatted text preview: c 1 is chosen so that r = a (when x = 0); therefore, c 1 = a . Also, the coecient c 2 must be chosen so that (at the right end of the truncated cone) we have r = b (when x = L ); therefore, c 2 = ( b a ) /L . Our expression, then, becomes r = a + b a L x . Substituting this into our previous statement and solving for the eld, we nd E = i a + b a L x 2 . Consequently, the potential dierence between the faces of the cone is V = Z L E dx = i Z L a + b a L x 2 dx = i L b a a + b a L x 1 L = i L b a 1 a 1 b = i L b a b a ab = iL ab . The resistance is therefore R = V i = L ab . (b) If b = a , then R = L/a 2 = L/A , where A = a 2 is the crosssectional area of the cylinder....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Current

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