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Unformatted text preview: 77. (a) From Gauss’ law, we get bg Er = c h 3
1 4 πρ r 3 r ρ r
4 πε 0 r 3
4 πε 0
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
ρr ( − ρ ) r − a
3ε 0 bg b g ...
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