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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.
 Spring '10
 Rebello,NobelS

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