ch23-p032 - 32. To evaluate the field using Gauss law, we...

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(b) Once outside the cylinder, Eq. 23-12 is obeyed. To find λ = q / L we must find the total charge q . Therefore, 0.04 21 1 0 1 2 1.0 10 C/m. q Ar rLdr LL π == × And the result, for r = 0.050 m, is 0 | | /2 3.6 N/C. Er λπε G 32. To evaluate the field using Gauss’ law, we employ a cylindrical surface of area 2 π r L where L is very large (large enough that contributions from the ends of the cylinder become irrelevant to the calculation). The volume within this surface is
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This note was uploaded on 09/07/2010 for the course PHYS 247 taught by Professor Zv during the Fall '10 term at The Petroleum Institute.

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