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17.59. Set Up:
The method of Example 17.10 shows that the electric field outside the sphere is the same as for a
point charge of the same charge located at the center of the sphere. The charge of an electron has magnitude
Solve: (a)
For
so
The number of excess electrons is
(b)
17.60. Set Up:
The charge distribution has spherical symmetry, so the electric field is radial and depends only on
the distance from the center of the charged sphere. The surface area of a sphere is
Solve: (a)
Apply Gauss’s law to a spherical surface of radius
concentric with the sphere of charge. The elec
tric field is constant over the Gaussian surface and perpendicular to it, so
The charge
enclosed is
Q,
so Gauss’s law gives
and
This is the same
E
as for a point charge
Q
at the
center of the sphere.
(b)
No direction is preferred, so
17.61. Set Up:
The charge distribution has spherical symmetry, so the electric field, if nonzero, is radial and
depends only on the distance from the center of the shell.
Solve: (a)
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 Spring '07
 Shoberg
 Charge

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