# UY1: Using Gauss’s Law For Common Charge Distributions

### Finding the electric field of an infinite line charge using Gauss’s Law

Electric charge is distributed uniformly along an infinitely long, thin wire. The charge per unit length is $\lambda$ (assumed positive). Find the electric field at a distance r from the wire.

First, we wrap the infinite line charge with a cylindrical Gaussian surface. We are unable to calculate the total charge enclosed by the cylindrical Gaussian surface. But we can analyze use Gauss’s Law using a per unit length approach.

We shall analyze a small portion (length $l$) of the infinite line charge. The charge enclosed will be $\lambda l$. Hence,

\begin{aligned} EA &= \frac{Q_{encl}}{\epsilon_{0}} \\ E \times 2 \pi r l &= \frac{\lambda l}{\epsilon_{0}} \\ E &= \frac{\lambda}{2 \pi \epsilon_{0} r} \end{aligned}

We’re done. Recall that we have found the electric field of an infinite line charge from the electric potential using Coulomb’s Law. In this case, Gauss’s Law is much faster.

What if it is not infinite? Can I still use Gauss’s Law to find the electric field?

No. (Technically you can. But it will be hard.) If it is not infinite, the electric field perpendicular to the two ends of the cylindrical Gaussian surface will not be zero. You will have to account for the electric flux at the two ends, which would not be easy. Hence, it is better to use Coulomb’s law.

### Finding the electric field of an infinite plane sheet of charge using Gauss’s Law

Find the electric field caused by a thin, flat, infinite sheet on which there is a uniform positive charge per unit area $\sigma$.

Imagine a cylindrical Gaussian surface covering an area A of the plane sheet as shown above. The charge enclosed will be: $\sigma A$.

Hence, (Note that the area enclosed is 2A as the electric field is coming out from both sides)

\begin{aligned} E \times \text{Area} &= \frac{\sigma A}{\epsilon_{0}} \\ E \times 2A &= \frac{\sigma A}{\epsilon_{0}} \\ E &= \frac{\sigma}{2 \epsilon_{0}} \end{aligned}

Notice that the electric field strength does not depend on the distance from the infinite sheet. As a comparison, recall that we calculated this electric field using Coulomb’s law as shown here.

### Finding the electric field between oppositely charged parallel infinite conducting plates using Gauss’s Law.

Two infinite plane parallel conducting plates are given charges of equal magnitude and opposite sign. The charge per unit length is $+ \sigma$ for one and $- \sigma$ for the other. Find the electric field in the region between the plates.

First, we observed that if we take a cube shaped Gaussian surface that encloses both conducting plates, there will be no electric field outside of the two plates. This is because the charge enclosed is 0 (the +ve and -ve charge cancels exactly), which gives 0 electric field by Gauss’s Law.

Then, we imagine a cube shaped Gaussian surface that only enclose one plate. Since we know that outside of the two plates have no electric field, the only electric field is in the region between the plates. The charge enclosed will be: $\sigma A$, where A is the area enclosed by the Gaussian surface that has electric field coming out/going in.

\begin{aligned} E \times \text{Area} &= \frac{Q_{encl}}{\epsilon_{0}} \\ E \times A &= \frac{\sigma A}{\epsilon_{0}}\\ E &= \frac{\sigma}{\epsilon_{0}} \end{aligned}

As a comparison, recall that we calculated this electric field using Coulomb’s law as shown here.

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