ch23_p87 - 87. (a) We note that the symbol “e” stands...

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Unformatted text preview: 87. (a) We note that the symbol “e” stands for the elementary charge in the manipulations below. From −e = ∞ 0 ρ ( r ) 4πr 2 dr = ∞ 0 3 A exp ( −2r / a0 )4πr 2 dr = πa0 A we get A = –e/πa03. (b) The magnitude of the field is E= = ( qencl 1 = e+ 2 2 4πε 0 a0 4πε 0 a0 5 e exp ( −2 ) 2 4πε 0 a0 a0 0 ) ρ (r )4πr 2 dr = . We note that E points radially outward. e 4πε a 2 0 0 1− 4 3 a0 a0 0 exp ( −2r / a0 ) r 2 dr ...
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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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