ch23-p022 - 22. We imagine a cylindrical Gaussian surface A...

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Here, the maximum value is ( ) () 8 4 max 12 2 2 0 2.0 10 C/m 1.2 10 N/C. 2 2 0.030 m 8.85 10 C / N m E r ε × λ == = × π π× 22. We imagine a cylindrical Gaussian surface A of radius r and unit length concentric with the metal tube. Then by symmetry enc 0 2. A q EdA rE ⋅= π = G G v (a) For r < R, q enc = 0, so E = 0. (b) For r > R, q
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This note was uploaded on 09/07/2010 for the course PHYS 268 taught by Professor Gn during the Fall '10 term at The Petroleum Institute.

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