Physics 21
Fall, 2012
Gauss’s Law
Here is a step-by-step procedure for working problems using
Gauss’s Law, which can be written
Qencl
E · dA =
.
(1)
0
S
1
Pick a gaussian surface S
The gaussian surface that you pick will have the same symmetry as the charge distribution that you are given. THERE
ARE ONLY THREE POSSIBILITIES.
• Sphere Use this when you expect the magnitude of the
electric field, E, to depend on the distance r from the
center of a spherical charge distribution.
2
Evaluate LHS of Eq. 1
If you chose your gaussian surface S wisely, the surface integral on the LHS of Gauss’s Law will be easy to evaluate.
If S is a sphere of radius r, then E is always perpendicular
to the surface of the sphere, and therefore always parallel to
dA, so
E · dA = 4πr2 E
S
If S is a cylinder of radius r and lenght l, then E will be
perpendicular to S on the curved part of the cylinder and
parallel to S on the flat base and cap of the cylinder, so the
only nonzero contribution to the surface integral comes from
the curved surface:
E · dA = 2πrlE
S
If S is a pillbox, then E will be parallel to S on the flat part
of the pillbox parallel to the planar charge distribution, and
perpendicular to E on the edges of the pillbox. The surface
integral will be
E · dA = EA
S
• Cylinder Use this when you expect E to depend on
r, which is now the distance from the central axis of a
cylindrical charge distribution.
The only exception to the rules above would be if your gaussian surface is entirely inside a conductor. In that case, use
the fact that E = 0 inside a conductor, so that
E · dA = 0.
S
3
Evaluate RHS of Eq. 1
Now evaluate Qencl , the charge enclosed by the gaussian surface. The answer will depend on whether the charge distribution is on an insulator or a conductor. (The charge on a
conductor will only be on the surface of the conductor; the
charge can be distributed uniformly throughout the volume
of an insulator.) Normally it just takes simple geometry to
work out Qencl ,
4
• Pillbox Use this when the charge distribution is planar,
i.e., the same for any values of the coordinates (x, y) in
a plane.
Equate your LHS and RHS
Once you have evaluated the LHS and RHS of Eq. 1, just
equate them and solve for the unknown. If you are given the
charge distribution and worked out Qencl , you will probably
solve for E. If you were asked to find the charge distribution
on a conductor, you may have used the fact that E = 0 inside
the conductor to argue that LHS = 0, and in this case you
will probably solve for the Qencl that makes the RHS equal
to zero.
You will find that any arbitrary dimensions that you chose
(length l of a cylinder or surface area A of the ends of a
pillbox) appear both on the LHS and RHS. These variables
will cancel.
October 8, 2012