Chapter 16 Gauss’ Law
If you want to find the center of mass of a potato, you can
do so by experiment or by laborious calculation, involving the
numerical evaluation of a triple integral. However, if the potato
happens to be a uniform ellipsoid, you know from its symmetry
exactly where the center of mass is without calculations. Such
are the advantages of symmetry. Symmetrical situations arise in
all areas of physics; when possible, it makes sense to cast the
laws of physics in forms that take full advantage of this fact.
Gauss’s law can be used to take advantage of special
symmetry situation. For electrostatics problems, it is the full
equivalent of Coulomb’s law.
Central to Gauss’ law is a hypothetical closed surface called
a Gaussian surface. The Gaussian surface can be of any shape
you wish to make it, but the most useful surface is one that
mimics the symmetry of the problem at hand. Thus the Gaussian
surface will often be a sphere, a cylinder, or some other
symmetrical form. It must be closed surface, so that a clear
distinction can be made between points that are inside the surface,
and outside the surface. Gauss’ law relates the electric fields at
points on a Gaussian surface and the net charge enclosed by that
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surface.
16.1 Flux of an Electric Field
1. To determine the flux of an
electric field, consider figure (a),
which
shows
an
arbitrary
Gaussian surface immersed in a
non-uniform electric field. Let us
divide the surface into small
squares of area A , each square
being small enough to permit us to neglect any curvature and
consider the individual square to be flat. We represent each

such element of area with an area vector A , whose

magnitude is the area A . Each vector A is perpendicular to
the Gaussian surface and directed away from the interior of the
surface.
2. Because the squares have been taken to be arbitrarily small,
the electric field

E
square. The vectors
may be taken as constant over any given

A

E
and
for each square then make
some angle  with each other. A provisional definition for the
flux of the electric field for the Gaussian surface of above
figure is
 
  E  A .
This equation instructs us to visit each
square on the Gaussian surface, to evaluate the scalar product
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 
E  A
for the two vectors

E
and

A
that we find there, and
to sum the results algebraically for all the squares that make
up the surface.
3. The exact definition of the flux of the electric field through a
closed surface is found by allowing the area of the squares
shown in above figure (a) to become smaller and smaller,
approaching a differential limit dA. The area vectors then
approach a differential limit

dA
. The sum of the above
equation then becomes an integral and we have, for the
definition of electric flux,
 
   E  dA .
The circle on the integral
sign indicates that the integration is to be taken over the entire
(closed) surface. The flux of the electric field is a scalar, and
its SI unit is the newton-square-meter per coulomb ( N  m 2 / C ) .
4. We can interpret above definition in the following way: First
recall that we can use the density of electric field lines passing
through an area as a measure of an electric field

E
there.
Specifically, the magnitude E is proportional to the number of
electric field lines per unit area. Thus, the dot product
 
E  dA
is
proportional to the number of electric field lines passing
through area

dA .
Then, because the integration is carried out
over a Gaussian surface, which is closed, we see that the
electric flux

through a Gaussian surface is proportional to
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the net number of electric field lines passing through that
surface.
16.2 Gauss’ Law
1. Gauss’ law relates the net flux

of an electric field through a
closed surface to the net charge qenc that is enclosed by that


 0    0  E  dA  q enc
surface. It tells us that
(Gauss' law) .
Above
equation hold only when the net charge is located in a vacuum
or in air. The net charge
is the algebraic sum of all the
q enc
enclosed positive and negative charges, and it can be positive,
negative, or zero. If
q enc
q enc
is positive, the net flux is outward; if
is negative, the flux is inward.
2. Charge outside the surface, no matter how large or how close
it may be, is not included in the term
q enc
in Gauss’ law. The
exact form or location of the charges inside the Gauss surface
is also of no concern; the
only things that matter are
the magnitude and sign of
the net enclosed charge.
3. The

E
on the left side,
however, is the electric field
resulting from all charges,
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both those inside and those outside the Gaussian surface.
4. See right figure.
16.3 A Charged Isolated Conductor
1. Gauss’ law permits us to prove an important theorem about
isolated conductors: if an excess charge is placed on an
isolated conductor, that amount of charge will move entirely
to the surface of the conduct. None of the excess charge will
be found within the body of the conductor. This might seem
reasonable, considering that charge with the same sign repel
each other. You might imagine that, by moving to the surface,
the added charge are getting as far away from each other as
they can. See Figure (a).
2. An isolated conductor
with a cavity: Figure (b)
shows
the
same
conductor, but now with
a cavity that is totally within the conductor. The distribution of
charge or the pattern of the electric field will not be charged
anymore.
3. The external electric field: The electric field

E
at and just
outside the conductor’s surface must be perpendicular to that
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surface, and its magnitude is
E

0
, in which  is the charge
per unit area. It means the magnitude of the electric field at a
location just outside a conductor is proportional to the surface
charge density at that location on the conductor. If the charge
on the conductor is positive, the electric field points away
from the conductor; it points toward the conductor if the
charge is negative.
16.4 Applying Gauss’ Law
1. Cylindrical symmetry
(1) Figure shows a section of an infinitely
long cylindrical plastic rod with a uniform
(positive) lines charge density  . The
expression for the magnitude of the electric field
distance r from the axis of the rod is
(2) The direction of

E
E

E
at a

.
20 r
is radially outward if the charge is
positive; and radially inward if it is negative.
2. Planer symmetry
(1) Figure shows a portion of a thin,
infinite, non-conducting sheet with a
uniform (positive) surface charge  .
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The magnitude of the electric field is
E

2 0
, and its
direction is perpendicular to and away from the sheet. Since
we are considering an infinite sheet with uniform charge
density, this result holds for any point at a finite distance from
the sheet.
(2) One conducting plate:
(3) Two conducting plates:
3. Spherical symmetry: (1) We can use Gauss’ law to prove the
two shell theorems: (a) A shell of uniform charge attracts or
repels a charged particle that is outside the shell as if all the
shell’s charge were concentrated at the center of the shell. (b)
A shell of uniform charge exerts no electrostatic force on a
charged particle that is located inside the shell.
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