ch23_p72 - 72. (a) From Gauss’ law, 3 1 qencl 1 4 πρr 3...

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Unformatted text preview: 72. (a) From Gauss’ law, 3 1 qencl 1 4 πρr 3 r ρr Er = r= = . 4 πε 0 r 3 4 πε 0 3ε 0 r3 bg c h (b) The charge distribution in this case is equivalent to that of a whole sphere of charge density ρ plus a smaller sphere of charge density –ρ which fills the void. By superposition Er = bg ρr ( − ρ ) r − a ρa + = . 3ε 0 3ε 0 3ε 0 b g ...
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.

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