Electric Field of Thick Infinite Sheet of Charge

[Ackbach]

It is known that a non-conducting sheet of charge (charge on only one "side") has electric field magnitude
$$E=\frac{\sigma}{2\varepsilon_0},$$
where $\sigma$ is the surface charge density in Coulombs per square meter. Suppose now that we have an infinite sheet, but it has a thickness $\ell$ to it and a uniform volume charge density $\rho$. The sheet is still a non-conductor, so that we CAN have a uniform volume charge density. The problem is to find the electric field at a point $P$ a distance $L$ away from the sheet.

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Assign a coordinate system as follows: the origin is on the surface opposite $P$, and is the closest point on that surface to $P$. Let $z$ be positive towards $P$, and let $x$ and $y$ be appropriately oriented in the surface opposite $P$ to produce a right-handed coordinate system. There is not sufficient symmetry, so far as I can tell, to use Gauss's Law, so I will perform the $dE=\dfrac{k \, dq}{r^2}$ integration.

[EDIT] Actually, if you argue by symmetry, and break up a Gaussian cylinder into 5 regions: the two endcaps, the lateral region inside the sheet, and the two lateral regions outside, there is sufficient symmetry to use Gauss's Law. The result is the same as below.

The chunk of charge $dq=\rho \, dV=\rho \, dx \, dy \, dz$. However, we will find it more convenient to use cylindrical coordinates $\langle s, \theta, z\rangle$, where $s,\theta$ are replacing $x,y$. Now, let
\begin{align*}
\mathbf{x}&=\text{vector from the origin to chunk of charge } dq =\langle x,y,z \rangle \\
\mathbf{r}&=\text{vector from chunk of charge } dq \text{ to }P =\langle -x, -y, \ell+L-z\rangle \\
\mathbf{R}'&=\text{vector from the origin to }P =\langle 0,0,\ell+L\rangle \\
\mathbf{R}&=\text{vector, from the projection of } \mathbf{x}\text{ onto } \mathbf{R}' \text{, to }P
=\langle 0,0,\ell+L-z\rangle.
\end{align*}
Note that $\mathbf{R}$ and $\mathbf{R}'$ point in the same direction - from the origin to $P$. Also, let $r=|\mathbf{r}|$ and $R=|\mathbf{R}|$.

Now then, we have that
\begin{align*}
dE&=\frac{k \, \rho \, dV}{r^2} \\
&=\frac{k \, \rho \, s \, ds \, d\theta \, dz}{r^2} \\
dE_z&=z \; \text{component of } dE =dE \, \cos(\varphi),
\end{align*}
where $\varphi$ is the angle between $\mathbf{R}$ and $\mathbf{r}$. Now,
$$\frac{\mathbf{r}\cdot\mathbf{R}}{rR}=\cos(\varphi),$$
so this quantity is exactly what we need. We have that
\begin{align*}
r&=\sqrt{s^2+(\ell+L-z)^2} \\
R&=\ell+L-z \\
\mathbf{r}\cdot\mathbf{R}&=(\ell+L-z)^2.
\end{align*}
Thus,
$$dE_z=\frac{k \, \rho \, s (\ell+L-z) \, ds \, d\theta \, dz}{[s^2+(\ell+L-z)^2]^{3/2}},$$
and
$$E_z=k\rho \int_0^{\ell}\int_0^{2\pi}\int_0^{\infty}\frac{s (\ell+L-z) \, ds \, d\theta \, dz}{[s^2+(\ell+L-z)^2]^{3/2}}
=2\pi k\rho \int_0^{\ell}\int_0^{\infty}\frac{s (\ell+L-z) \, ds \, dz}{[s^2+(\ell+L-z)^2]^{3/2}}.$$
The substitution $t=\ell+L-z$ reduces these integrals down to
$$E_z=2\pi k\rho \int_0^{\infty}s\int_L^{\ell+L}\frac{t \, dt \, ds}{[s^2+t^2]^{3/2}},$$
where I have swapped the order of integration. The inside integral is perfectly tractable using a $u$ substitution, although for some reason Wolfram Programming Cloud chokes on it. I get that
$$\int_L^{\ell+L}\frac{t \, dt}{[s^2+t^2]^{3/2}}=\frac{1}{\sqrt{s^2+L^2}}-\frac{1}{\sqrt{s^2+(\ell+L)^2}},$$
making the overall integral
$$E_z=2\pi k\rho \int_0^{\infty}\left[\frac{s}{\sqrt{s^2+L^2}}-\frac{s}{\sqrt{s^2+(\ell+L)^2}}\right] ds.$$
Now this integral Wolfram Programming Cloud can handle, and it yields a final result of simply $\ell$. Therefore,
$$E_z=2\pi k\rho \ell,$$
which essentially is the same result as before, if you think formally of $\sigma=\rho \ell$. But this is surprising, is it not?

Consider the original result, which was for an infinitesimal sheet of charge. If you add up a finite thickness of such infinitesimal sheets, wouldn't you expect the final result to be infinite? That is, in thickness $\ell$ there are uncountably infinitely many sheets of charge - or you can think of it this way. Why the finite result in my derivation, then? Did I make a mistake?

Many thanks if you slogged through this entire post!

 

[I like Serena]

Hey Ackbach,

I think there is sufficient symmetry to apply Gauss's law.
Due to symmetry, the electric field will be uniform and perpendicular to the sheet.
Suppose we pick a block around the sheet that extends to some distance on both sides, say distance $L$, and that intersects the sheet over an area $A$, then:
$$\int \frac{\rho_{free}}{\varepsilon_0} dV = \oint \mathbf D \cdot d\mathbf S
\qquad\Rightarrow\qquad \frac{\rho \ell A}{\varepsilon_0} = 2DA
\qquad\Rightarrow\qquad D = \frac{\rho\ell}{2\varepsilon_0}$$

I'm using $\mathbf D$ here because the material is not a conductor.
For a homogeneous, isotropic, nondispersive, linear material, we have the relationship $\mathbf D = \varepsilon\mathbf E$.

 

[Ackbach]

Hey Ackbach,

I think there is sufficient symmetry to apply Gauss's law.
Due to symmetry, the electric field will be uniform and perpendicular to the sheet.
Suppose we pick a block around the sheet that extends to some distance on both sides, say distance $L$, and that intersects the sheet over an area $A$, then:
$$\int \frac\rho{\epsilon_0} dV = \oint \mathbf D \cdot d\mathbf S
\qquad\Rightarrow\qquad \frac{\rho \ell A}{\epsilon_0} = 2DA
\qquad\Rightarrow\qquad D = \frac{\rho\ell}{2\epsilon_0}$$

Yeah, I just figured that out myself as well, using cylinders (see the edit in the OP). But it still doesn't address the question of adding up an infinite number of infinitesimal thicknesses, each contributing a finite amount, and getting a finite total!

 

[I like Serena]

Yeah, I just figured that out myself as well, using cylinders (see the edit in the OP). But it still doesn't address the question of adding up an infinite number of infinitesimal thicknesses, each contributing a finite amount, and getting a finite total!

Doesn't each sheet of infinitesimal thickness contain an infinitesimal charge, and therefore contributes an infinitesimal amount?

 

[Ackbach]

Doesn't each sheet of infinitesimal thickness contain an infinitesimal charge, and therefore contributes an infinitesimal amount?

It may have an infinitesimal amount of charge, but it contributes a finite amount: $E=\sigma/(2\varepsilon_0)$ is the field at $P$ due to one infinitesimally thin sheet of charge. That's a finite, real number. The issue, it seems to me, is one of dimensions, and of balancing infinities. For an infinitesimal sheet, you have an infinitesimal thickness, but infinite length and infinite width.

It occurred to me that you might consider what would happen to the thick sheet if you took a rectangular prism of it, with finite breadth and width, and depth $\ell$, and condensed it to the surface near $P$. That is, you're essentially converting a volume charge density to a surface charge density. This is sort of what would happen if, all of a sudden, you changed the entire slab from a non-conductor to a conductor. Then what would the surface charge density be? Well, the total charge in this finite (not even infinitesimal!) chunk is $\rho V=\rho A \ell$. But now, distributed only over the surface, the surface charge density would be $\sigma=\rho A \ell/A=\rho \ell$, like I mentioned in the OP.

This wouldn't be a valid move at all, except that infinite sheets of charge don't care how far away you are! So, from this argument, you can see how the finite result makes sense. So, on the one hand, I have the explicit calculation and Gauss's Law, both directly and indirectly as I have just outlined, all pointing to a finite field. On the other hand, I have a mathematical argument that says it ought to be infinite. At this point, I am doubting the validity of the mathematical argument. But if so, there must be a fallacy or mistaken assumption in it somewhere. What is that fallacy or mistaken assumption?

 

[Ackbach]

Actually, each infinitesimal sheet contains an infinite amount of charge. If the charge density on a sheet is $\sigma$, then the charge in any finite area would be $\sigma A$. If the area is infinite, as it is for the infinite sheet, then the charge would be infinite. The trick is that the sheet gets infinitely far away from the field point.

I'm thinking about my mathematical argument, and I'm wondering if there isn't some basic mistake I'm making there. Consider an infinitesimal sheet of charge, surface charge density $\sigma$. Then its field is $E=\sigma/(2\varepsilon_0)$. How should we add up all fields due to all the sheets between $z=0$ and $z=\ell$?

Normally, if I have a function like $y=x^2$, then to find a differential, I would do $dy=2x \, dx$. The problem with the above equation is that the RHS is a constant. Technically, then, if I took the differential of the LHS, I should get zero on the RHS. That is, $dE=0$. Or am I overthinking it? Should I just do
\begin{align*}
E&=\frac{\sigma}{2\varepsilon_0} \\
\int_0^{\ell}E \, dz&=\int_0^{\ell}\frac{\sigma}{2\varepsilon_0} \, dz \\
E\ell&=\frac{\sigma \ell}{2\varepsilon_0}.
\end{align*}

But then I get back to where I started. I'm confused!

 

[I like Serena]

I believe the contribution is \(\displaystyle \frac{\rho d\ell}{2\varepsilon_0}\), which is an infinitesimal amount, instead of \(\displaystyle \frac{\sigma}{2\varepsilon_0}\)
An infinitesimal thin sheet doesn't have a finite non-zero surface charge density $\sigma$ - it has density $d\sigma = \rho d\ell$.
If we integrate it over the thickness $\ell$, it sums up to the surface charge density $\sigma$.