## The Electric Field from an Infinite Line Charge

This second walk through extends the application of Gauss's law to an infinite line of charge. This time cylindrical symmetry underpins the explanation. Once again interactive text, visualizations, and mathematics provide a rich and easily understood presentation.

When we worked with a point charge we recognized spherical symmetry, which inspired us to select a spherical surface to simplify Gauss's law. Can we find a similar symmetry for an infinite line charge? If we do a little bit of experimenting with the charge and field line diagram, we see that rotation around the axis of the charged line does not change the shape of the field. This leads to a Gaussian surface that curves around the line charge. Since Gauss's law requires a closed surface, the ends of this surface must be closed. This is a cylinder. This symmetry is commonly referred to as cylindrical symmetry.

The field lines are everywhere perpendicular to the walls of the cylinder, and they are evenly distributed around the surface. Just as with the sphere that surrounded a point charge. And like that sphere, $\mathbf{E}\cdot d\mathbf{A}$ will be constant over the surface.

The ends of the cylinder will be parallel to the electric field so that $\mathbf{E}\cdot d\mathbf{A}=0$ , and once again Gauss's law will be simplified by the choice of surface.

#### An Infinite Line Charge Surrounded By A Gaussian Cylinder

Exploit the cylindrical symmetry of the charged line to select a surface that simplifies Gausses Law.

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### Gauss's Law

As always start with Gauss's Law.

$${\iint}_{S}\mathbf{E}\cdot d\mathbf{A}=4\pi q$$We break the surface integral into three parts for the left cap, $L$ , the body, $B$ , and the right cap, $R$ , of the cylinder.

$${\iint}_{S}\mathbf{E}\cdot d\mathbf{A}={\iint}_{L}\mathbf{E}\cdot d\mathbf{A}+{\iint}_{B}\mathbf{E}\cdot d\mathbf{A}+{\iint}_{R}\mathbf{E}\cdot d\mathbf{A}$$Let's work with the left end cap, $L$ , first.

$${\iint}_{L}\mathbf{E}\cdot d\mathbf{A}={\iint}_{L}E\mathrm{cos}\left(\theta \right)dA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is parallel to the cap.}={\iint}_{L}E\mathrm{cos}\left(\frac{\pi}{2}\right)dA=0$$This makes a great deal of sense. When the field is parallel to the surface the flux through the surface is zero. The process is identical for the right cap, so all the contributions to the flux come from the body of the cylinder.

$${\iint}_{B}\mathbf{E}\cdot d\mathbf{A}={\iint}_{B}E\mathrm{cos}\left(\theta \right)dA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is perpendicular to the body.}={\iint}_{B}E\mathrm{cos}\left(0\right)dA={\iint}_{B}EdA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is constant over the surface.}=E{\iint}_{B}dA=EA\phantom{\rule{0ex}{0ex}}\text{If}L\text{is the length of the cylinder.}=E2\pi rL$$Now that we have the flux through the cylinder wall, we need the right side of the equation, the charge contained within the surface. Since $\lambda $ is the charge density of the line the charge contained within the cylinder is:

$$4\pi q\phantom{\rule{3.5em}{0ex}}=4\pi \lambda L$$Setting the two haves of Gauss's law equal to one another gives the electric field from a line charge as

$$E\phantom{\rule{4.4em}{0ex}}=\frac{2\lambda}{r}$$Then for our configuration, a cylinder with radius $r=15.00\text{cm}$ centered around a line with charge density $\lambda =8\raisebox{1ex}{$\text{statC}$}\!\left/ \!\raisebox{-1ex}{$\text{cm}$}\right.$

$$E\phantom{\rule{4.4em}{0ex}}=\frac{2\lambda}{r}=\frac{2\left(8\raisebox{1ex}{$\text{statC}$}\!\left/ \!\raisebox{-1ex}{$\text{cm}$}\right.\right)}{15.00\text{cm}}=1.07\frac{\text{statV}}{\text{cm}}$$Next we build on this to find the electric field from a charged plane.