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# Electrostatics: Of Special Interest to Physics C Students

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By McGraw-Hill Professional
Updated on Feb 12, 2011

Practice problems for these concepts can be found at:

Electrostatics Practice Problems for AP Physics B & C

A more thorough understanding of electric fields comes from Gauss's law. But before looking at Gauss's law itself, it is necessary to understand the concept of electric flux.

The electric flux, ΦE, equals the electric field multiplied by the surface area through which the field penetrates.

Flux only exists if the electric field lines penetrate straight through a surface. (Or, if the electric field lines have a component that's perpendicular to a surface.) If an electric field exists parallel to a surface, there is zero flux through that surface. One way to think about this is to imagine that electric field lines are like arrows, and the surface you're considering is like an archer's bulls eye. There would be flux if the arrows hit the target; but if the archer is standing at a right angle to the target (so that his arrows zoom right on past the target without even nicking it) there's no flux.

In words, Gauss's law states that the net electric flux through a closed surface is equal to the charge enclosed divided by ε0. This is often written as

### How and When to Use Gauss's Law

Gauss's law is valid the universe over. However, in most cases Gauss's law is not in any way useful—no 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!

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 4πR2. So the charge on the conductor is σR2. Now, the Gaussian surface of radius r has area 4πr2. Plug all of this into Gauss's law:

E · 4πr2 = σR2/ε0.

All the variables are known, so just solve for electric field: E = σR20r2.

Now we can state our answer, where r is the distance from the center of the charged sphere:

What is interesting about this result? We solved in terms of the charge density σ on the conductor. If we solve instead in terms of Q, the total charge on the conductor, we recover the formula for the electric field of a point charge:

Now you see why the Coulomb's law constant k is the same thing as

Practice problems for these concepts can be found at:

Electrostatics Practice Problems for AP Physics B & C

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