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Homework 1 Answers, 95.657 Electromagnetic Theory I
Dr. Christopher S. Baird, UMass Lowell
Jackson 1.1
Use Gauss's theorem
∮
S
E
⋅
n
da
=
q
0
and
∮
E
⋅
d
l
=
0
to prove the following:
a) Any excess charge placed on a conductor must lie entirely on its surface. (A conductor by definition
contains charges capable of moving freely under the action of applied electric fields.)
b) A closed, hollow conductor shields its interior from fields due to charges outside, but does not shield
its exterior from fields due to charges placed inside it.
c) The electric field at the surface of the conductor is normal to the surface and has a magnitude
/
0
, where
is the charge density per unit area on the surface.
SOLUTION:
a) First, the problem contains the unstated assumption that what is wanted is the location of the charges
in
static equilibrium
. This assumption is valid because this problem is found in the chapter on
electrostatics. Static equilibrium (the lack of movement) can only exist when free charges are present if
there are no electric fields.
Therefore the electric field inside a conductor is zero.
One can draw any
arbitrary, closed surface completely inside the conductor and it will always have zero electric field at
every point on the surface. The integral of the electric field over this surface is zero and by Gauss's law,
the total charge is therefore zero. Because the surface is arbitrary, it can be chosen to be infinitesimally
small or be chosen to go through any point in the conductor's interior. Therefore every point inside a
conductor has zero charge, and all the charge must then reside on the surface.
b) Assume for the moment that the electric field external to the conductor is perpendicular to the
conductor's
surface at the surface (the proof is left to part c). Because the conductor is hollow, the
interior region of the conductor has by definition no free charges. If we choose a closed mathematical
surface just inside and parallel to the inner surface of the hollow conductor, it therefore contains a total
charge of zero. Gauss's Law then becomes:
∮
S
E
⋅
n
da
=
0
Because the mathematical surface of the integral is parallel to the surface of the conductor, and the
electric field is perpendicular to the surface of the conductor, the electric field must be parallel to the
normal:
This simplifies Gauss's law to:
E
n
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S
E da
=
0
The magnitude of a vector, in this case
E
,
is always positive. There is no way to get the integral of a
permanently positive function to equal zero except if the function itself is zero at every point. Thus the
electric field is zero at every point on the mathematical surface. The surface can be chosen to show that
all points in the hollow region have zero electric fields.
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 Spring '11
 Avery
 Charge, Magnetism, Mass, Work

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