ch23-p030 - 30. (a) In Eq. 23-12, = q/L where q is the net...

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(b) Since the field is zero inside the conductor (in an electrostatic configuration), then there resides on the inner surface charge –q , and on the outer surface, charge +q (where q is the charge on the rod at the center). Therefore, with r i = 0.05 m, the surface density of charge is 9 92 inner 2.0 10 C/m 6.4 10 C/m 2 2 2 (0.050 m) ii q rL r σ π −λ × == = = × ππ for the inner surface. (c) With r o = 0.10 m, the surface charge density of the outer surface is outer
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This note was uploaded on 09/07/2010 for the course PHYS 221 taught by Professor Hk during the Fall '10 term at The Petroleum Institute.

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