P24_018 - E max = 2 πrε =(2 × 10 − 8 C m 2 π(0 030...

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18. We imagine a cylindrical Gaussian surface A of radius r and unit length concentric with the metal tube. Then by symmetry I A ~ E · d ~ A =2 πrE = q enclosed ε 0 . (a) For r>R , q enclosed = λ ,so E ( r )= λ/ 2 πrε 0 . (b) For r<R , q enclosed =0,so E = 0. The plot of E vs r is shown below. Here, the maximum value is
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Unformatted text preview: E max = λ 2 πrε = (2 . × 10 − 8 C / m) 2 π (0 . 030 m) (8 . 85 × 10 − 12 C 2 / N · m 2 ) = 1 . 2 × 10 4 N / C . 10000 E 0.02 0.04 0.06 r...
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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.

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