ch23-p077 - 77. (a) From Gauss’ law, we get bg Er = c h 3...

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Unformatted text preview: 77. (a) From Gauss’ law, we get bg Er = c h 3 1 qencl 1 4 πρ r 3 r ρ r r= = . 4 πε 0 r 3 4 πε 0 r3 3ε 0 (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 ρr ( − ρ ) r − a ρa Er = + = . 3ε 0 3ε 0 3ε 0 bg b g ...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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