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Problem-Solving Strategy
Gauss’s Law
IDENTIFY the relevant concepts: Gauss’s law is most useful in situations where
the charge distribution has spherical or cylindrical symmetry or is distributed
uniformly over a plane. In these situations we determine the direction of
from
the symmetry of the charge distribution. If we are given the charge distribution, we
can use Gauss’s law to find the magnitude of
Alternatively, if we are given the
field, we can use Gauss’s law to determine the details of the charge distribution. In
either case, begin your analysis by asking the question, “What is the symmetry?”
SET UP the problem using the following steps:
1.
Select the surface that you will use with Gauss’s law. We often call it a
Gaussian surface. If you are trying to find the field at a particular point, then
that point must lie on your Gaussian surface.
2.
The Gaussian surface does not have to be a real physical surface, such
as a surface of a solid body. Often the appropriate surface is an imaginary
geometric surface; it may be in empty space, embedded in a solid body, or both.
3.
Usually you can evaluate the integral in Gauss’s law (without using a
computer) only if the Gaussian surface and the charge distribution have some
symmetry property. If the charge distribution has cylindrical or spherical
symmetry, choose the Gaussian surface to be a coaxial cylinder or a concentric
sphere, respectively.
EXECUTE the solution as follows:
1.
Carry out the integral in Eq. (22.9). This may look like a daunting task,
but the symmetry of the charge distribution and your careful choice of a
Gaussian surface makes it straightforward.
2.
Often you can think of the closed Gaussian surface as being made up of
several separate surfaces, such as the sides and ends of a cylinder. The integral
over the entire closed surface is always equal to the sum of the
integrals over all the separate surfaces. Some of these integrals may be zero, as
in points 4 and 5 below.
3.
If
is perpendicular (normal) at every point to a surface with area A,
if it points outward from the interior of the surface, and if it also has the same
magnitude at every point on the surface, then
over that surface is equal to EA. If instead
inward, then
and
and
is perpendicular and
4.
If
is tangent to a surface at every point, then
integral over that surface is zero.
5.
If
and the
at every point on a surface, the integral is zero.
6.
In the integral
is always the perpendicular component
of the total electric field at each point on the closed Gaussian surface. In
general, this field may be caused partly by charges within the surface and partly
by charges outside it. Even when there is no charge within the surface, the field
at points on the Gaussian surface is not necessarily zero. In that case, however,
the integral over the Gaussian surface—that is, the total electric flux through the
Gaussian surface—is always zero.
7.
Once you have evaluated the integral, use Eq. (22.9) to solve for your
target variable.
EVALUATE your answer: Often your result will be a function that describes how
the magnitude of the electric field varies with position. Examine this function with a
critical eye to see whether it makes sense.