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(This might be somewhat related to a previous question I posted here, however it seemed different enough to warrant a separate post.)

I'm trying to see how, for a point charge at the origin, I might apply the differential form of Gauss' Law to integrate and find the electric field, for example along the $x$-axis:

$$\nabla \cdot E= \frac{dE}{dx}=\frac{\rho}{\epsilon}$$

Essentially I'm trying to follow the procedure depicted on slide 13 of this lecture except for the case of a point charge at the origin. My understanding is that such a point charge would need to be represented by a Dirac delta charge density at the origin, which would make sense to me. However, integrating this Dirac delta would seem to produce a constant electric field rather than an electric field that approaches infinity at the origin and falls off as $\frac{1}{x^2}$. I'm confused overall how this decreasing electric field would be predicted as Gauss' law seems to suggest there would need to be negative charge density at those locations, whereas the charge density is zero almost everywhere.

In my previous question linked above my error came from trying to use a noncontinuous expression for the electric field which did not have accurate partial derivatives, however here it seems that the charge density should definitely be 0 everywhere except the origin so I'm not sure what the fix might be.

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  • $\begingroup$ I need to think about it but.... The reduction from 3D (where you can find the field of a point charge or any symmetrical charge distribution) pretty easily from the integral form of Gauss' equation, to 1D, does not come without a price. For instance, what is the field for $x > x_0$ in your slide #13? Can you calculate it? In other words, you can't find back the $\dfrac{1}{x^2}$ law from the 1D problem, because you neglected what happens in the other two dimensions. But I need to fully put toghether an explanation as to why this is. $\endgroup$ Mar 28, 2023 at 21:06

2 Answers 2

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$\rho = \sigma \delta(x)$ describes an infinite charged sheet with charge density $\sigma$, not a point charge. As is well known and as you have found out, this produces an electric field that is constant on either side of the sheet.

To describe a point charge $q$ at the origin, you would write the charge density as $\rho = q\delta(x)\delta(y)\delta(z)$ in Cartesian coordinates, or $\rho = \frac{q}{2\pi r^2}\delta(r)$ in spherical coordinates. The problem is not one dimensional in Cartesian coordinates. It is one dimensional in spherical coordinates, and you will obtain the inverse square law if you use the divergence operator in spherical coordinates.

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  • $\begingroup$ Thanks a lot the answer, this is very helpful. So in that first case, since $y, z$ aren't specified, the charge distribution is nonzero for all points $(0,y,z)$ leading to the infinitely thin and wide charge sheet? And additionally, in the slide 13 I referenced in the original post, is the charge actually implicitly infinite in extent in the y and z dimensions? $\endgroup$
    – Halleff
    Mar 29, 2023 at 2:53
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    $\begingroup$ @knzy Yes, $\delta(x)$ means that the charge density diverges at $(0, y, z)$. Yes, that is the implicit assumption on slide 13, although in practice it can just mean the charge extends far enough in y and z to be considered infinite in extent. $\endgroup$
    – Puk
    Mar 29, 2023 at 5:53
  • $\begingroup$ Thank you- another question on that second one, I'm wondering how to understand the solution for $E$ due to the uniform charge density as on slide 13. For $x>x_1$, the electric field is constant, which makes sense to me since thereafter all the charge looks similar to the thin charged sheet with a constant electric field. But for $x<0$ the field is constant at a different value (depending on the choice of the additive constant), which seems to somehow break symmetry. It’s true that $\rho$ isn’t symmetric about 0, but this coordinate choice shouldn’t affect the physical solution, no? $\endgroup$
    – Halleff
    Mar 29, 2023 at 16:52
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    $\begingroup$ The field doesn't have to be symmetric, precisely because there is a boundary condition only on the left side and not on the right side. It's not true that the situation can never be symmetric for any boundary condition: consider the condition $E(x = 0) = -\sigma/2\epsilon_0$, where $\sigma = \int_0^{x_1}\rho(x) dx$ is the charge per unit area. Otherwise your description of the solution correct and I'm not sure where your confusion is. $\endgroup$
    – Puk
    Mar 29, 2023 at 21:15
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    $\begingroup$ Ah okay, that cleared it up. For some reason I simply forgot that a symmetric result would involve E being equal in magnitude but opposite in direction on either side of the charge block's midpoint (just like the charged sheet) rather than equal in both mag/direction. That result is perfectly intuitive to me. Thanks a lot for your help! $\endgroup$
    – Halleff
    Mar 29, 2023 at 21:38
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So, as promised, I thought about your question a little longer and I'm coming with this provisional answer.

As Puk showed, the actual point charge problem/distribution is inherently 3D. Also, his first sentence actually explains it all:

ρ=σδ(x) describes an infinite charged sheet with charge density σ, not a point charge

Acually this single sentence sums it all, however since I did not realize it when I first read it, I am expanding a bit, in case you're like me :)

The lessons you quoted are geared towards basic semiconductor electronics, in which you often use this 1-D approximation of Poisson's equation (or Gauss' equation, if you're interested in the E field).

The underlying assumption in this type of development is that you have an infinitely extended crystal lattice, in both directions of a certain plane. The only thing that changes, can change in the remaining (perpendicular) direction.

For instance, when you consider the classic pn junction (or an MOS structure, it's the same idea), you bring together two crystal materials but are only interested in what happens along the direction where the material changes from p to n:

Image Courtesy: Wikipedia

Although the "bars" of material are shown to be finite, actually the underlying idea is that they are extending indefinitely in all directions which are not $x$.

So, when you are considering only $x$, and notice the density distribution $\rho(x)$, what you're actually implying is that this same distribution extends indefinitely in all directions but $x$.

Which is to say, as Puk pointed out:

ρ=σδ(x) describes an infinite charged sheet with charge density σ, not a point charge

Then, it should be clear that what you get out of that distribution can never be the field of a point charge. You actually get the field of a charge sheet, which is linearly growing inside the portion with the charge density, and constant afterwards. Like in a classic capacitor.

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