P24_046 - 46. (a) From Gauss’ law, E (r ) = 1 qencl 1...

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Unformatted text preview: 46. (a) From Gauss’ law, E (r ) = 1 qencl 1 (4πρr3 /3)r ρr r= = . 4πε0 r3 4πε0 r3 3 ε0 (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 E (r ) = (−ρ)(r − a) ρa ρr + = . 3 ε0 3 ε0 3 ε0 ...
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