Electric Field of Thick Infinite Sheet of Charge

In summary, the conversation discusses the electric field magnitude of a non-conducting sheet of charge with charge on only one "side". It is found to be equal to $E=\frac{\sigma}{2\varepsilon_0}$, where $\sigma$ is the surface charge density in Coulombs per square meter. The conversation then considers an infinite sheet with a thickness $\ell$ and a uniform volume charge density $\rho$, and aims to find the electric field at a point $P$ a distance $L$ away from the sheet. It is determined that there is sufficient symmetry to use Gauss's Law, and the result is found to be the same as using the integration method. The conversation also addresses the question of how adding up
  • #1
Ackbach
Gold Member
MHB
4,155
89
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.

---

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!
 
Mathematics news on Phys.org
  • #2
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 $\mathbf D = \varepsilon\mathbf E$.
 
  • #3
I like Serena said:
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!
 
  • #4
Ackbach said:
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?
 
  • #5
I like Serena said:
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?
 
  • #6
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!
 
  • #7
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 $\sigma$.
 

Related to Electric Field of Thick Infinite Sheet of Charge

1. What is an electric field?

An electric field is a physical quantity that describes the strength and direction of the force experienced by a charged particle in the presence of other charged particles. It is a vector quantity, meaning it has both magnitude and direction.

2. What is a thick infinite sheet of charge?

A thick infinite sheet of charge is a hypothetical model used in electromagnetism to represent a flat surface with a uniform distribution of charge. It is considered "infinite" because it extends infinitely in all directions, and "thick" because it has a finite thickness.

3. How is the electric field of a thick infinite sheet of charge calculated?

The electric field of a thick infinite sheet of charge can be calculated using the formula E = σ/2ε0, where σ is the surface charge density (charge per unit area) and ε0 is the permittivity of free space. This formula holds true for points above, below, or on the surface of the sheet.

4. What is the direction of the electric field for a thick infinite sheet of charge?

The electric field of a thick infinite sheet of charge is always perpendicular to the surface of the sheet. This means that the direction of the electric field will either be pointing away from the sheet (if the sheet has a positive charge) or towards the sheet (if the sheet has a negative charge).

5. How does the magnitude of the electric field change as you move further away from the sheet?

For a thick infinite sheet of charge, the magnitude of the electric field decreases as you move further away from the sheet. This decrease follows an inverse-square law, meaning that the electric field is proportional to 1/r2, where r is the distance from the sheet. This means that the electric field will be stronger closer to the sheet and weaker further away from it.

Similar threads

Replies
2
Views
1K
Replies
21
Views
2K
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
5
Views
1K
Replies
1
Views
822
Replies
1
Views
827
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
865
Replies
1
Views
830
  • Atomic and Condensed Matter
Replies
4
Views
2K
Back
Top