one expects you to be able to evaluate a three-dimensional integral with a dot product! ONLY use Gauss’s
law when the problem has spherical, cylindrical, or planar symmetry.
First, identify the symmetry of the problem. Then draw a closed surface, called a “Gaussian surface,”
that the electric field is everywhere pointing straight through. A Gaussian surface isn’t anything real ...
it’s just an imaginary closed surface that you’ll use to solve the problem. The net electric flux is just E
times the area of the Gaussian surface you drew.
You should NEVER, ever, try to evaluate the integral ∫ E · dA in using Gauss’s law!Here is an example problem.
Consider a metal sphere of radius R that carries a surface charge density σ . What is the magnitude of
the electric field as a function of the distance from the center of the sphere?There are two possibilities here. One possibility is that the function describing the electric field will be a
smooth, continuous function. The other possibility is that the function inside the sphere will be different
from the function outside the sphere (after all, they’re different environments—inside the sphere you’re
surrounded by charge, and outside the sphere you’re not). So we’ll assume that the function is different in
each environment, and we’ll consider the problem in two parts: inside the sphere and outside the sphere.
If it turns out that the function is actually smooth and continuous, then we’ll have done some extra work,
but we’ll still get the right answer.
Inside the sphere, draw a Gaussian sphere of any radius. No charge is enclosed, because in a
conductor, all the charges repel each other until all charge resides on the outer edge. So, by Gauss’s law
since the enclosed charge is zero, the term E·A has to be zero as well. A refers to the area of the Gaussian
surface you drew, which sure as heck has a surface area.
The electric field inside the conducting sphere must be zero everywhere . This is actually a general
result that you should memorize—the electric field inside a conductor is always zero.
Outside the sphere, draw a Gaussian sphere of radius r . This sphere, whatever its radius, always
encloses the full charge of the conductor. What is that charge? Well, σ represents the charge per area of
the conductor, and the area of the conductor is 4πR 2 . So the charge on the conductor is σ 4 πR 2 . Now, the
Gaussian surface of radius r has area 4πr 2 . Plug all of this into Gauss’s law:
E ·4πr 2 = σ 4 πR 2 /ε 0 .All the variables are known, so just solve for electric field: E = σR 2 /ε 0 r 2 .
Now we can state our answer, where r is the distance from the center of the charged sphere: