- #1

Ackbach

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$$E=\frac{\sigma}{2\varepsilon_0},$$

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!