Is the charge distribution for an electric field unique?

[Ganesh Ujwal]

If the electric field and boundary conditions are known exactly for a region of space, is it true that there exists only one charge distribution in that region of space that could have produced it?

My understanding of the uniqueness theorem in electrostatics is that for a given charge distribution and boundary conditions for a volume, there exists only one (unique) solution to Poisson's equation, and thus the electric field in that volume is uniquely determined. Does the arrow point the other way, too? If we know the field and boundary conditions, is the charge distribution uniquely determined in the volume? Is there a simple example that illustrates why or why not?

[studiot]

No, but the converse is true.

Classically, there is only one field configuration generated by a specific charge distribution, within a specified region of space.

 

Is this a question about classical physics or otherwise?

 

If you know and understand Poisson's and Laplace's equations, you should have come across Gauss's Flux Law, which provides any number of simple examples.

 

Is this homework?

 

 

I would also like to point out that as Laplace's and Poisson's equations are partial differential equations, their general solutuons contain arbitrary functions so cannot be unique.

Particular solutions are extracted by means of the boundary conditions.

Edited December 18, 2014 by studiot

[timo]

Gauss Law tells you the field defines the total charge in the enclosed volume. This is true for any sub-volume in your field. Hence, for any region the field defines the charge, including arbitrary small regions. Hence, knowing the field defines the charges.

 

Only exception is if you count a negative and a positive charge at the same position to be different than having no charge at this location (or 2x plus and 1x minus to be different than a single plus charge). This can not be resolved - neither by the argument with Gauss Law nor by the the field at all, because only the net charge density goes into the field in the first place.

Edited December 18, 2014 by timo

[studiot]

 

Gauss Law tells you the field defines the total charge in the enclosed volume. This is true for any sub-volume in your field. Hence, for any region the field defines the charge, including arbitrary small regions. Hence, knowing the field defines the charges. Only exception is if you count a negative and a positive charge at the same position to be different than having no charge at this location (or 2x plus and 1x minus to be different than a single plus charge). This can not be resolved - neither by the argument with Gauss Law nor by the the field at all, because only the net charge density goes into the field in the first place.

 

 

It would be interesting if you were to elaborate and confirm whether you think the answer to the OP is yes or no.

[timo]

I self-conceived my post as a rather obvious "yes, the field defines the charge distribution". I'll add a paragraph to make it more clear.

[swansont]

I self-conceived my post as a rather obvious "yes, the field defines the charge distribution". I'll add a paragraph to make it more clear.

 

 

But it doesn't. e.g. If you have a known field outside of a conducting shell, all you know is the enclosed net charge. The distribution of that charge inside the shell is arbitrary. An infinite number of possibilities.

[timo]

I was assuming the case that the volumes in which you know the field and the volume whose charge you ask about are identical (which could be inferred from the OP asking about "in that region of space"). You are indeed correct that knowing the field in region A does not automatically tell you the charge distribution in a different region B.

[studiot]

Aw swansont, you spoilt my surprise.

 

Take any (isolated) point, P, in space.

Let the Field region be specified as the space between a sphere centered on that point of radius R and infinity.

 

For any charge Q the field is the same whether it is located at P or on any sphere, also centered at P with radius < R .

 

I was careful before because I don't know if the OP has to find an example or counterexample for homework.

Edited December 18, 2014 by studiot

[swansont]

The OP is a tad vague with regard to what you actually know.

[timo]

The OP seems very clear to me with respect to identical regions for field and charges. But you two guys seem to have a different opinion than me, so I accept this as experimental proof. I guess the best thing is to wait and see what the OP makes out of the replies generated so far.

 

In the meantime, there is another issue I just happened to think about (warning to OP: this probably goes beyond what you were actually asking): At least in some rather pathological cases you can get into a situation where you can not determine the charges. Take a 1D world with periodic boundary conditions and a uniform charge density, for example. For symmetry reasons there should not be any non-zero vector-valued field and the potential must be constant. I think you cannot tell a zero charge density from a non-zero one in this case. Not really sure what that means. But what is actually bugging me: If there is one instance where knowing the field in region A does not tell you the charge in region A (despite the charge simply being the derivative), what other instances are there?

[studiot]

Another situation is embodied in the method of (virtual) images, much beloved of textbooks for illustations and exercises in plotting fields. Textbooks usually tonly treat symmetrical situations, but the mehtod has application in the practical world in the estimation of work functions.

Edited December 18, 2014 by studiot

[swansont]

what other instances are there?

 

 

2D case with an infinite sheet of charge — it gives you a uniform field. The charges could be at the edge of your defined space or outside of it, and could be + charges on one side or - charges on the other, or both.

[timo]

Not sure I am getting this one. The scenario I had in mind was having the field in a volume V and deducing the charge distribution in V. From what I understand your example still is about not being able to take the field in V and deduce the charges outside V (let's assume the volume to be an open volume, meaning to behave like an open interval, for simplicity).

 

My line of thinking with the 1D example (which works analogously in higher dimensions) goes towards that you can only find deviations from the vacuum state and that in the example the respective vacuum state is the uniform charge density. But it leaves the taste of having flung around a buzzword (vacuum state) to create the illusion of explanation without being any step further to understand what goes wrong, there.

Edited December 18, 2014 by timo

[studiot]

 

If the electric field and boundary conditions are known exactly for a region of space, is it true that there exists only one charge distribution in that region of space that could have produced it?

 

 

The answer is still No, even if the OP thinks he wishes to consider only the charges within the region since it is entirely possible to have zero charges within the region and the entire field and boundary conditions produced by an external charge distribution, even if there is an 'internal' charge distribution that could produce the specified conditions.

Edited December 19, 2014 by studiot

[timo]

Given that I sketched an explicit algorithm of how to deduce the charge density from the field in my first post I don't think that "the answer is still no" really adds value to the discussion. A counter-example or an explanation why Gauss law does not apply would be great. I can also rephrase my first post if you have trouble understanding it (adding the paragraph apparently did not help).

 

Consider a differential equation dE/dx = d. Knowing d in a certain interval does not define E, since there is an integration constant that can be chosen arbitrarily. This integration constant can be fixed by a boundary condition, as the OP already mentioned. Knowing E, on the other hand, uniquely defines d. It can be calculated by means of taking the derivative of E with respect to x (because of d = dE/dx and the derivative being unique). Now, imagine E to be the electric field and d to be the charge density.

[swansont]

Not sure I am getting this one. The scenario I had in mind was having the field in a volume V and deducing the charge distribution in V. From what I understand your example still is about not being able to take the field in V and deduce the charges outside V (let's assume the volume to be an open volume, meaning to behave like an open interval, for simplicity).

 

My line of thinking with the 1D example (which works analogously in higher dimensions) goes towards that you can only find deviations from the vacuum state and that in the example the respective vacuum state is the uniform charge density. But it leaves the taste of having flung around a buzzword (vacuum state) to create the illusion of explanation without being any step further to understand what goes wrong, there.

 

The OP does not restrict the case to being an open interval. It mentions boundary conditions, which to me implies a closed interval.

 

If you have a cube with a uniform field in the x direction, what is the charge distribution that created it?

[timo]

The charge distribution in the cube is [math] \nabla \vec E = \nabla \text{const} = 0[/math]. The charge distribution outside the cube that created the field is not the question here, as I mentioned in pretty much every post in this thread.

Edited December 19, 2014 by timo

[studiot]

 

Given that I sketched an explicit algorithm of how to deduce the charge density from the field in my first post I don't think that "the answer is still no" really adds value to the discussion. A counter-example or an explanation why Gauss law does not apply would be great. I can also rephrase my first post if you have trouble understanding it (adding the paragraph apparently did not help).

 

I'm sorry. how did this add value to the discussion?

 

Here is a field plot.

 

Prove the charge distribution to be unique.

 

I apologise in advance if you have more sophisticated drawing facilities than I do.

 

 

 

[timo]

Here is a field plot. Prove the charge distribution to be unique.

 

 

I do not completely get your drawing. I feel you still talk about an example where knowing the field in the volume outside a sphere does not tell you the charge distribution inside a sphere. This is absolutely correct: "inside" and "outside" are different volumes, even if one is topologically enclosed in the other.

 

In any other case two possible attempts of a proof would be:

 

Proof 1:

Gauss Law tells you the field defines the total charge in the enclosed volume. This is true for any sub-volume in your field. Hence, for any region the field defines the charge, including arbitrary small regions. Hence, knowing the field defines the charges.

 

 

Proof 2:

The charge distribution is given by [math] \rho = \nabla \vec E[/math]. The derivative is unique. Therefore, so is the charge distribution. It's of course the same as in proof 1, but some people are more impressed by mathematical symbols than by construction algorithms.

 

 

I have the feeling I am repeating myself a bit too often to still consider this a discussion. Looks more like a battle of egos to me. So I am out of this thread. Feel free to learn something from what I have said here or ignore it.

[studiot]

Ego has nothing to do with it.

 

 

Gauss Law tells you the field defines the total charge in the enclosed volume.

 

Gauss' Law does no such thing, for every conceivable volume.

 

 

I feel you still talk about an example where knowing the field in the volume outside a sphere

 

Nor is this about feeling. It is strictly about science.

 

I have never seen a geometric sphere with corners. Can you show me one?

 

Anyone looking at my drawing will see a dashed square, clearly labelled 'section'.

This section can only be part of a 3 dimensional volume.

 

Equally the only place where a field is shown in my sketch is within that volume.

 

You claimed, and I asked you to demonstrate, that you can deduce the charges that produce that specified field, from the field alone.

 

If you were unclear about anything in the sketch the proper course of action was to ask for an explanation - I would have been happy to oblige.

Edited December 19, 2014 by studiot

[swansont]

 

 

The charge distribution in the cube is [math] \nabla \vec E = \nabla \text{const} = 0[/math]. The charge distribution outside the cube that created the field is not the question here, as I mentioned in pretty much every post in this thread.

 

 

The surface is not outside of the volume.

[elfmotat]

The disagreement here seems to stem from whether or not the complete electric field [math]\mathbf{E} (\mathbf{r} )[/math] is known, or just the field at some specific points. I don't think anyone would disagree that complete knowledge of [math]\mathbf{E} (\mathbf{r} )[/math] ensures complete knowledge of [math]\rho (\mathbf{r})[/math], by the fact that [math] \epsilon_0 \nabla \cdot \mathbf{E} (\mathbf{r} ) = \rho (\mathbf{r})[/math].

[studiot]

 

The disagreement here seems to stem from whether or not the complete electric field is known, or just the field at some specific points. I don't think anyone would disagree that complete knowledge of ensures complete knowledge of , by the fact that .

 

 

Thank you for trying to intercede, helpfully.

 

But, as has already been pointed out, the specified existence of boundary conditions requires boundaries.

Edited December 19, 2014 by studiot