# Electric field of infinite sheet

I have a question regarding the electric field of an infinite sheet in the $$x-y$$-plane at $$z=0$$ with a constant sheet charge density $$\sigma$$. There are many ways to obtain the result $$\boldsymbol E=\frac{\sigma}{2\epsilon_0}\frac{z}{|z|}\boldsymbol{e_z}$$ but I can't seem to recreate this result using the divergence theorem and a cylindric volume. Maybe someone can point out an error in my thoughts: Let's assume first $$\boldsymbol E=E(z)\boldsymbol{e_z}$$. Then we put a cylindric volume $$V$$ into the sheet such that it gets cut in half and its symmetry axis is the $$z$$-axis. Furthermore I want this cylinder to have height $$2z$$ (so that the top is located at $$z$$ and the bottom is located at $$-z$$) and radius $$R$$. Now, by the divergence theorem we have $$\begin{equation}\int_V\text{div}\boldsymbol EdV=\int_{\partial V}\boldsymbol E\cdot\boldsymbol{dA}.\tag{1}\end{equation}$$ By the maxwell equation $$\text{div}\boldsymbol E=\frac{\sigma}{\varepsilon_0}$$ we easily see that the left-hand side of $$(1)$$ is equal to $$\frac{\sigma}{\epsilon_0}\int_VdV=\frac{\sigma}{\epsilon}V=\frac{\sigma}{\epsilon_0}\pi R^2\cdot 2z.$$ Now for the right-hand side we first need the surface elements:

• $$\boldsymbol{dA}=Rd\varphi dz\boldsymbol{e_\varphi}$$ for the lateral surface,
• $$\boldsymbol{dA}=zrdrd\varphi\boldsymbol{e_z}$$ for the top and
• $$\boldsymbol{dA}=-zrdrd\varphi(\boldsymbol{-e_z})$$ for the bottom.

Maybe this is already where my error is since this now means that the lateral surface cancels ($$\boldsymbol{e_\varphi}\perp\boldsymbol{e_z}$$) and we have $$\int_{\partial V}\boldsymbol E\cdot\boldsymbol{dA}=2zE(z)\int_0^R\int_0^{2\pi}rd\varphi dr=2\pi R^2 E(z)z$$ So here I must have lost a factor of 2 and also I don't obtain the fact that the electric field changes the direction at $$z=0$$. Whereas my second problem can probably be fixed by just assuming $$\boldsymbol E(z)=\begin{cases}E(z)\boldsymbol{e_z},& z> 0\\ -E(z)\boldsymbol{e_z}, &z<0\end{cases},$$ I have no idea how to fix the first problem: Where did my $$2$$ go?

The issue is in your definition of density, turns out that the volume density of charge in your problem is

$$\rho = \sigma\delta(z - 0)= \sigma \delta(z) \tag{1}$$

That way Maxwell's equation becomes

$$\require{cancel} \int_{\partial V}{\rm d}^2{\bf S}\cdot {\bf E} = \int_V{\rm d}^3{\bf r}~\nabla \cdot {\bf E} = \int_V{\rm d}^3{\bf r} \frac{\rho}{\epsilon_0} = \frac{\sigma}{\epsilon_0} \int{\rm d}x{\rm dy}\cancelto{1}{\int{\rm d}z \delta(z)} = \frac{\sigma A}{\epsilon_0} \tag{2}$$

where $$A$$ is the area of your cylinder $$A = \pi R^2$$. Now, on the l.h.s you are right, split the boundary into three pieces, the normal vector on the sides is perpendicular to the field and its contribution will vanish. Leading to

$$\int_{\partial V}{\rm d}^2{\bf S}\cdot {\bf E} = 2E A \tag{3}$$

Putting these two results together

$$E = \frac{\sigma }{2\epsilon_0}$$

• Thanks for your answer! However, taking the rong density of charge cannot be my only mistake. Because my r-h-s in (1) depends on $z$. So what is the problem with my surface element? Maybe it is the $\pm z$ I tossed in there? But if I don't do that then the top and bottom integral cancel each other out as well. – RedLantern Dec 2 '18 at 7:34
• @RedLantern The surface element is also wrong, it should be ${\rm d}A = r{\rm d}r d\phi$. If you multiply that by ${\rm d}z$ (as you have it), you will get a volume differential – caverac Dec 2 '18 at 8:17
• Yes, I thought about this. But the surface element on the top points to the positive $z$-axis where the surface element on the bottom points towards the negative $z$-axis. So if we bring the direction $\boldsymbol{e_z}$ into the element (which we have to), then one is exactly the multiple of $-1$ of the other. This means that adding up the integrals yields $0$. Do you understand where my problem is? – RedLantern Dec 2 '18 at 8:54
• @RedLantern The field lines point up $+z$ in the upper face, and down in the lower one, so the inner product is alway positive. No need to introduce an extra factor ${\rm d}z$ into the differential – caverac Dec 2 '18 at 8:58
• I never introduced a $dz$ in my differential. But I understand what you're saying. So we have to implement the direction of the electric field right away, it doesn't come naturally by computing those integrals. – RedLantern Dec 2 '18 at 9:23

Your mistake comes from a misunderstanding about the charge density. It is not uniformly distributed through the volume of the cylinder; rather, it is concentrated in an infinitesimally thin sheet; namely, the volumetric charge density (which is what Gauss's Law gives you) is $$\rho=\sigma\delta(z)$$. So your integral on the left-hand side of (1) should be

$$\frac{\rho}{\epsilon_0}\int dV = \frac{\sigma}{\epsilon_0}\int dA \int \delta(z) dz$$

Since, for any region of integration containing $$0$$, we have that $$\int \delta(z) dz =1$$, therefore, this integral evaluates to

$$\frac{\sigma}{\epsilon_0}\pi R^2$$

which is the total charge enclosed by your Gaussian surface divided by $$\epsilon_0$$. Incidentally, this would have been completely obvious had you used the integral form of Gauss's Law instead of the differential form.

• Thanks for your answer too! Since you both basically described the same problem I replied above. – RedLantern Dec 2 '18 at 7:35