## The Electric Field From Parallel Charged Planes

This fourth and final Gauss's law example examines two parallel charged planes. Previous examples include a point charge, a line charge, and a charged plane. The two charged plane example is longer, but only slightly more complex. No new symmetry is introduced, rather the decomposition of a problem into smaller simpler pieces is stressed.

This time we tackle two parallel oppositely charged planes. We build directly on the charged plane example, including more logic and physical reasoning, that is more physics. The new physics guides a slightly different application and interpretation of the same mathematics as with the single charged plane. Keep in mind the difference between length and complexity. This problem is, since it involves two planes, twice the size of the single plane example. However, it is not nearly twice as complex. Decomposing large problems into smaller problems is a common and powerful approach. Many seemingly complex problems are really a set of smaller simpler problems in disguise.

#### Overall Approach

We get a running start by recalling the translational symmetry from the charged plane. As we move parallel to the infinite planes every point looks like every other point, with two planes extending infinitely in every direction. Just as before, this requires that the field be constant at a constant distance from the planes, and that the field is perpendicular to the planes.

To find the field both between and outside of the planes, we will use two Gaussian surfaces and some thought. The first Gaussian surface will span both planes and will determine the electric field outside of the planes. The second Gaussian surface will span only one plane and will determine the field between the planes.

Reflection was a useful symmetry when we worked with a single charged plane. Now we have an asymmetric configuration, so things are different. The asymmetry leads us to expect fields that appear opposite to one another when viewed from each side. For example, an inward bound field toward the negative side, and an outward bound field from the positive side.

Looking at the planes from a few cm to the right, we see a positively charged plane and a negatively charged plane stretching to infinity. The positive plane is closer, so any field is expected to be directed outward from the positive plane.

From a few cm to the left of the planes, we see a negatively charged plane and a positively charged plane stretching to infinity. From the left, the negative plane is closer and we expect any field to be directed towards the negative plane.

#### Parallel Charged Planes

Two parallel oppositely charged planes. Our Gaussian surface can enclose no charge, the charge from one plane, or charge from both planes.

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#### The First Surface

Here we determine the field outside of the planes. Evaluate Gauss's law for a cylinder spanning both planes with one cap to the left, one cap to the right, and the body connecting them.

$${\iint}_{S}\mathbf{E}\cdot d\mathbf{A}\phantom{\rule{.6em}{0ex}}=4\pi q$$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.

$${\iint}_{S}\mathbf{E}\cdot d\mathbf{A}\phantom{\rule{.6em}{0ex}}={\iint}_{L}{\mathbf{E}}_{L}\cdot d\mathbf{A}+{\iint}_{B}\mathbf{E}\cdot d\mathbf{A}+{\iint}_{R}{\mathbf{E}}_{R}\cdot d\mathbf{A}$$Let's work with the left end cap, $L$ , first.

$${\iint}_{L}{\mathbf{E}}_{L}\cdot d\mathbf{A}={\iint}_{L}{E}_{L}\mathrm{cos}\left(\theta \right)dA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is inward bound.}={\iint}_{L}{E}_{L}\mathrm{cos}\left(\mathrm{\pi}\right)dA={\iint}_{L}-{E}_{L}dA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is constant over the surface.}=-{E}_{L}{\iint}_{L}dA=-{E}_{L}A=-{E}_{L}\pi {r}^{2}$$We attach a subscript, $L$ , to the the electric field as we no longer have a convenient symmetry to tell us the relationship between the left and right fields so we keep track of them separately.

The contribution from the right side cap is computed the same way, and produces almost the same result. Except for the right cap, $\theta =0$ and ${E}_{R}$ will not gain a $-$ sign. So the total contribution from both caps is

$${\iint}_{L}{\mathbf{E}}_{L}\cdot d\mathbf{A}+{\iint}_{R}{\mathbf{E}}_{R}\cdot d\mathbf{A}=\left({E}_{R}-{E}_{L}\right)\pi {r}^{2}$$For the body of the cylinder, the key is to see that the field is parallel to the surface, and no field lines cross the surface.

$${\iint}_{B}\mathbf{E}\cdot d\mathbf{A}\phantom{\rule{.6em}{0ex}}={\iint}_{B}E\mathrm{cos}\left(\theta \right)dA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is parallel to the body.}={\iint}_{B}E\mathrm{cos}\left(\frac{\pi}{2}\right)dA=0$$Adding up all the contributions to get the total flux

$${\iint}_{L}{\mathbf{E}}_{L}\cdot d\mathbf{A}+{\iint}_{B}\mathbf{E}\cdot d\mathbf{A}+{\iint}_{R}{\mathbf{E}}_{R}\cdot d\mathbf{A}=-{E}_{L}\pi {r}^{2}+0+{E}_{R}\pi {r}^{2}=\left({E}_{R}-{E}_{L}\right)\pi {r}^{2}$$Now we can look at the right side of Gauss's law, the enclosed charge. The planes are oppositely charged so that the left plane charge density is the opposite of the right plane charge density, ${\sigma}_{R}=-{\sigma}_{L}=\sigma $

$$4\pi q\phantom{\rule{3.75em}{0ex}}=4\pi \left({\int}_{L}{\sigma}_{L}dA+{\int}_{R}{\sigma}_{R}dA\right)\phantom{\rule{0ex}{0ex}}\sigma \phantom{\rule{.5em}{0ex}}\text{is constant over each area.}=4\pi \left({\sigma}_{L}{\int}_{L}dA+{\sigma}_{R}{\int}_{R}dA\right)=4\pi \left({\sigma}_{L}\pi {r}^{2}+{\sigma}_{R}\pi {r}^{2}\right)\phantom{\rule{0ex}{0ex}}\text{We know that}{\sigma}_{L}=-{\sigma}_{R}\text{.}=4\pi \left(-{\sigma}_{R}+{\sigma}_{R}\right)\pi {r}^{2}=0$$Setting the two parts of Gauss's law equal to one another now gives us a constraint on the fields rather than directly giving a value as before.

$${E}_{R}-{E}_{L}\phantom{\rule{1.0em}{0ex}}=0$$This is where it gets interesting. Gauss's law can not directly give us a value for the field outside of the planes. But it does give us a lot to work with. The field is independent of position, so it is independent of distance from the planes. Also, the field on the right has the same magnitude and direction as the field on the left. We should expect this from the translational symmetry and the asymmetry under reflection.

Ask yourself, what does the field look like from an infinite distance? If we are looking back on the planes from an infinite distance, what do we see? As our distance approaches infinity the small distance separating the planes fades to insignificance. They appear as a single plane with the combined charge. The combined charge is zero, so at a very large distance we expect the electric field to be zero. Combine this with the fact that Gauss's law tells us the field is constant leads to a zero field everywhere outside of the charged planes because it is zero at a large distance.

#### The Second Surface

Now that we know the field outside of the planes, we can find the field between the planes. Shift the right cap of the Gaussian cylinder so that it is between the planes while keeping the left cap fixed. This leaves the left cap in a region of zero field and puts the right cap in the field between the planes. We show that the only contribution to the flux integral in Gauss's law is from the right cap and this will allow us to find the field between the planes. As is the usual case with a Gaussian cylinder, we break the flux integral into three parts.

$${\iint}_{S}\mathbf{E}\cdot d\mathbf{A}\phantom{\rule{.6em}{0ex}}={\iint}_{L}{\mathbf{E}}_{L}\cdot d\mathbf{A}+{\iint}_{B}\mathbf{E}\cdot d\mathbf{A}+{\iint}_{R}{\mathbf{E}}_{R}\cdot d\mathbf{A}$$Once again start with the left end cap, $L$ . This is particularly quick because ${\mathbf{E}}_{L}$ is $0$ from the first part.

$${\iint}_{L}{\mathbf{E}}_{L}\cdot d\mathbf{A}=0$$Next we can tackle the flux through the body of the cylinder. Here the flux is zero because the field is along the body of the cylinder.

$${\iint}_{B}\mathbf{E}\cdot d\mathbf{A}\phantom{\rule{.6em}{0ex}}={\iint}_{B}E\mathrm{cos}\left(\theta \right)dA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is parallel to the body.}={\iint}_{B}E\mathrm{cos}\left(\frac{\pi}{2}\right)dA=0$$The only non zero contribution to the flux will be from the right cylinder cap. Luckily we have already done the calculation in the first half of the problem. This time though, the field will be directed from the positive plane towards the negative plane.

$${\iint}_{R}{\mathbf{E}}_{R}\cdot d\mathbf{A}={\iint}_{R}{E}_{R}\mathrm{cos}\left(\theta \right)dA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is inward bound.}={\iint}_{R}{E}_{R}\mathrm{cos}\left(\mathrm{\pi}\right)dA={\iint}_{R}-{E}_{R}dA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is constant over the surface.}=-{E}_{R}{\iint}_{R}dA=-{E}_{R}A=-{E}_{R}\pi {r}^{2}$$This is the only contribution to the flux, so we move on to the enclosed charge. Again this is very similar to an earlier calculation. Remember that we kept the left end of the cylinder fixed and shifted the right end between the planes. Only the left plane is spanned by the cylinder.

$$4\pi q\phantom{\rule{3.75em}{0ex}}=4\pi {\int}_{L}{\sigma}_{L}dA\phantom{\rule{0ex}{0ex}}\sigma \phantom{\rule{.5em}{0ex}}\text{is constant.}=4\pi {\sigma}_{L}{\int}_{L}dA=4\pi {\sigma}_{L}\pi {r}^{2}$$Setting these expressions for the flux and charge equal to one another, and canceling common terms gives for the field between the plates

$$E\phantom{\rule{4.4em}{0ex}}=4\pi \sigma $$The field outside of the planes is zero, and the field between the planes is twice that from a single plane.

#### Summary

The left cylinder cap is in an area of zero field, and the body is parallel to the field, neither contributes to the flux through the surface. All flux contributions come from the right cap. We have positioned the cylinder so that the field is constant and perpendicular to the cap, so the flux is $EA$ where $A$ is the area of the cap.

The contained charge is the sum of the charge density of the planes, if any, intersected by the cylinder multiplied by the area, $A$ .

Setting the flux and the contained charge equal to each other, and dropping $A$ , which is common to both expressions gives

$$E\phantom{\rule{4.5em}{0ex}}=4\pi \left(\sum {\sigma}_{i}\right)$$For our configuration, with a charge density of ${\sigma}_{R}=-{\sigma}_{L}=.1\raisebox{1ex}{$\text{statC}$}\!\left/ \!\raisebox{-1ex}{${\text{cm}}^{2}$}\right.$ , we have at $x=70.0$