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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?

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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

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

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  • $\begingroup$ 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. $\endgroup$ – RedLantern Dec 2 '18 at 7:34
  • $\begingroup$ @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 $\endgroup$ – caverac Dec 2 '18 at 8:17
  • $\begingroup$ 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? $\endgroup$ – RedLantern Dec 2 '18 at 8:54
  • $\begingroup$ @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 $\endgroup$ – caverac Dec 2 '18 at 8:58
  • $\begingroup$ 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. $\endgroup$ – RedLantern Dec 2 '18 at 9:23
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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.

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

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