# Gauss law with two infinite planes

The electric field of an infinite plane above its surface is $$E=\frac{\rho}{2\epsilon_0}$$, where $$\rho$$ is the surface charge density and $$\epsilon$$ is the permittivity of free space. If we have two planes, I know that we can find the total electric field by superposition.

My question is, given two planes of two different charge density, say $$\rho$$ for the bottom one and $$-2\rho$$ for the top one. Obviously, the field in between the planes is perpendicular to them.

But using Gauss law, if we enclose only one of the surfaces, we get that the electric field between the planes should only depend on the charge density of that plane, but by superposition we have that $$E=\frac{3\rho}{2\epsilon_0}$$, which is not equal to any one field generated by one plane.

What went wrong in my application of Gauss law?

In the original application of Gauss's law to one infinite plate of charge, you have 2 Gaussian surfaces over which you have to integrate - the two faces of the pill box parallel to the plate (the 4 faces perpendicular to the plate you don't have to integrate over because the E-field will be parallel to those surfaces and so the flux will be 0). You can assume that these 2 faces have the same electric fields moving across them (you set up the Gaussian surface so that these 2 faces are equidistant from the plate) and so you get the standard result for electric field due to an infinite plate.

When you have 2 infinite plates, you can no longer say that the 2 faces have the same electric fields moving across them since you can't have the two faces being equidistant from both plates at the same time. So, Gauss's law is still valid, but you just can't factor out the electric field in the equation like you could with one plate. The easy way to solve the problem then is by using superposition.

Gauss's law is always correct, you just need a high degree of symmetry to easily apply it. If the symmetry is broken, the law becomes harder to apply.

EDIT: As per request: Gauss's law says that the total integral over a closed surface of the electric flux is equal to the enclosed charge. That's all it says. For charges which are not enclosed by the Gaussian surface, they don't affect this total integral because any electric flux going into the gaussian surface will also exit out of that Gaussian surface. Gauss's law doesn't say that external charges don't affect the electric field itself on that surface.

• From the OP's question, I believe the OP is also assuming the E-field described in Gauss' Law is the E-field resultant only from the enclosed charge i.e. not the "final" E-field. Cf "if we enclose only one of the surfaces, we get that the electric field between the planes should only depend on the charge density of that plane" Perhaps add something in the explanation to that effect for this OP and others? – Paul Oct 19 '18 at 18:17
• Actually the OP is mathematically right in their application of Gauss's law. If you apply it you get a true expression since the flux from the plate not in the Gaussian surface doesn't contribute to the integral (as you have said). So it's less of a matter of Gauss's law being "harder to apply" here, and more of how you can't even have a Gaussian surface that encloses both plates and has a relevant surface between them. The crux of the answer that is relevant to the OP is actually in the edited part. – Aaron Stevens Oct 20 '18 at 11:40
• So actually Gauss law become less useful when you have multiple charges, as it can't easily get you the total field. Thanks! – Shamaz Oct 20 '18 at 15:17

So you are actually correct in your math, just not in the interpretation of Gauss's law. I will first write out the math you have explained to make sure we are all on the same page.

Let's call the field from the positive plate $$\mathbf {E}_1$$, and the field from the negative plate $$\mathbf {E}_2$$. Let's put our Gaussian surface on the plate with a charge density $$\rho$$. Then on the surface between the plates the total field is $$E_1\hat x + E_2\hat x$$, and on the surface outside of the plates (closer to plate 1) the total field is $$-E_1\hat x + E_2\hat x$$

Now let's use Gauss's law, where we only need to worry about the surfaces mentioned above due to symmetry. $$\frac {Q_{enc}}{\epsilon_0}=\int (E_1\hat x + E_2\hat x)\cdot(dA\hat x)+\int(-E_1\hat x + E_2\hat x)\cdot(-dA\hat x)$$

The integrals are very easy to evaluate.$$^*$$ If we do this and divide by the area of the plate enclosed by our surface we get. $$\frac{\rho}{\epsilon_0}=E_1+E_2+E_1-E_2=2E_1$$ which is what you discuss in your answer. We end up with a true expression, but not the expression we wanted. Why is, this?

Notice what happens here: the $$E_2$$ terms cancel. This is because the field due to any charges outside our Gaussian surface does not contribute any net flux. It does not mean that the field between the plates itself is not influenced by the second plate, which can be seen by the fact that $$E_2$$ stuck around until it cancelled after calculating the total flux.

What you would really want is a Gaussian surface that contains "an equal amount" of both plates , but also has a surface between the plates that the total field flows through. I can't think of a surface with this property. So the easier way to go is treat each plate separately and then add the fields to get the total field.

$$^*$$ This is where my answer differs from the one by @enumaris. You actually can factor the fields out of each integral, since they are still uniform over the surfaces in question.

• It actually makes a lot of sense. So actually in a problem where you have multiple charges, to get the total field you must find a surface that enclose all of them and is also highly symmetric. Thanks! – Shamaz Oct 20 '18 at 15:15
• @Shamaz Yes. Single spheres, lines, and planes are the ideal geometries to apply Gauss's law to get the electric field. – Aaron Stevens Oct 20 '18 at 18:54

When deriving the formula for an infinite sheet of charge, we can consider a Cube as the Gaussian surface and thus

$$\int_{back} \vec{E}\cdot \vec{dA} + \int_{front} \vec{E}\cdot \vec{dA}= \dfrac{q_{enc}}{\epsilon_0}$$ ,

we set them both equal due to symmetery and thus arrive at $$\vec{E}=\dfrac{\sigma}{2\epsilon_0}$$.

In your case of two plates the back and front will be constant but will not be identical.