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Unformatted text preview: 37. The ﬁeld is radially outward and takes on equal magnitudevalues over the surface of any sphere centered
at the atom’s center. We take the Gaussian surface to be such a sphere (of radius r). If E is the magnitude
of the ﬁeld, then the total ﬂux through the Gaussian sphere is Φ = 4πr2 E . The charge enclosed by the
Gaussian surface is the positive charge at the center of the atom plus that portion of the negative charge
within the surface. Since the negative charge is uniformly distributed throughout the large sphere of
radius R, we can compute the charge inside the Gaussian sphere using a ratio of volumes. That is, the
negative charge inside is −Zer3 /R3 . Thus, the total charge enclosed is Ze − Zer3 /R3 for r ≤ R. Gauss’
law now leads to
r3
r
Ze
1
4πε0 r2 E = Ze 1 − 3
−3.
=⇒ E =
2
R
4πε0 r
R ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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