## The Electric Field from an Infinite Charged Plane

The exploration of Gauss's law continues with an infinite charged plane. Translational symmetry illuminates the path through Gauss's law to the electric field. Deeply interactive content visualizes and demonstrates the physics.

In our previous examples a rotational symmetry inspired us to choose a curved surface to simplify Gauss's law. With an infinite plane we have a new type of symmetry, translational symmetry. A moment's thought convinces us that if we move parallel to the plane, then any point looks like any other point. The plane always extends infinitely in every direction. As a result, we expect the field to be constant at a constant distance from the plane. This same symmetry tells us the field will be perpendicular to the plane. If the field had a component parallel to the plane, it would imply that there is something special about that direction. But for an infinite plane there is nothing special about any direction, so any component of the field along the plane would be a contradiction.

Another symmetry, reflection, allows us to say that the field is equal in magnitude, but opposite in direction on the other side of the plane. Looking towards the plane from a few cm away, I see an infinite wall of positive charge, and expect a field directed away from the wall.

You, looking at the plane from a few cm away on the opposite side see exactly the same thing. You see an infinite wall of positive charge, and expect a field directed away from the wall.

In the earlier examples we sought a surface with some parts that were parallel to the field, and other parts that were perpendicular to it. We can do the same thing here. Start by choosing surfaces perpendicular to the electric field on opposite sides of the charged plane. Obvious candidates have easy to compute surface areas, like a circle or a square. Traditionally, a circle is used. We place two circular surfaces a distance $h$ from each side of the plane. A Gaussian surface must be closed, so we connect to two surfaces, and we have a cylinder.

#### A Charged Plane With A Gaussian Pillbox

The electric field from the plane passes perpendicularly through the ends of the Gaussian pillbox and does not pass through the sides at all.

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### Gauss's Law

Once again we tackle Gauss's law. You have seen it so often, you can probably recite it from memory now.

$${\iint}_{S}\mathbf{E}\cdot d\mathbf{A}=4\pi q$$We reuse the technique from the line charge and break the surface integral into three parts: the left cap, the body, and the right cap.

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

$${\iint}_{L}\mathbf{E}\cdot d\mathbf{A}={\iint}_{L}E\mathrm{cos}\left(\theta \right)dA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is perpendicular to the cap.}={\iint}_{L}E\mathrm{cos}\left(0\right)dA={\iint}_{L}EdA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is constant over the surface.}=E{\iint}_{L}dA=EA=E\pi {r}^{2}$$This fits the pattern we have seen before. When the electric field is constant and perpendicular to the surface, the flux is $EA$ .

The contribution from the right side cap will be computed in the same way, and produce the same result. So the total contribution from both caps is

$${\iint}_{L}\mathbf{E}\cdot d\mathbf{A}+{\iint}_{R}\mathbf{E}\cdot d\mathbf{A}=2E\pi {r}^{2}$$For the body of the cylinder, the key is to see that the field is parallel to the surface.

$${\iint}_{B}\mathbf{E}\cdot d\mathbf{A}={\iint}_{B}E\mathrm{cos}\left(\theta \right)dA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is parallel to the body.}={\iint}_{B}E\mathrm{cos}\left(\frac{\pi}{2}\right)dA=0$$Adding up all the contributions

$${\iint}_{L}\mathbf{E}\cdot d\mathbf{A}+{\iint}_{B}\mathbf{E}\cdot d\mathbf{A}+{\iint}_{R}\mathbf{E}\cdot d\mathbf{A}=E\pi {r}^{2}+0+E\pi {r}^{2}=2E\pi {r}^{2}$$So the electric flux is twice the field strength times the area of the cylinder cap. This fits nicely with the idea that the field is outward bound, and hence contributes $EA$ , at each cap.

To extract useful information from Gauss's law we also need the enclosed charge, which is just the charge density of the plane, $\sigma $ , times the cross sectional area where the surface intersects with the plane.

$$4\pi q\phantom{\rule{3.5em}{0ex}}=4\pi \sigma \pi {r}^{2}$$It is important that the cross sectional area of the cylinder appears on both sides of Gauss's law. Canceling common terms from the last two equations gives the electric field from an infinite plane

$$E\phantom{\rule{4.4em}{0ex}}=2\pi \sigma $$This is independent of position!

For our configuration, with a charge density of $\sigma =.30\raisebox{1ex}{$\text{statC}$}\!\left/ \!\raisebox{-1ex}{${\text{cm}}^{2}$}\right.$ , we have

$$E\phantom{\rule{4.4em}{0ex}}=2\pi \left(.30\raisebox{1ex}{$\text{statC}$}\!\left/ \!\raisebox{-1ex}{${\text{cm}}^{2}$}\right.\right)=1.88\frac{\text{statV}}{\text{cm}}$$Next up we will look at the application of Gauss's law to a system of charges, parallel charged planes.