Understanding Gauss's Law: Where Does the Argument Break Down?

[TeethWhitener]

As already stated, this is not what is done when going from the integral form of Gauss’s law to the differential form. You only need the divergence theorem for compact volumes to do that.

I'm not sure I understand. I thought this was the divergence theorem for compact volumes.

 

[Metmann]

The 1/r^2 Green's function pre-assumes that the field of the point charge vanishes at infinity. The integrals you will obtain from this will clearly not be convergent (working in the potential now, which is usually much more convenient than working with the fields). This is just the point you noticed about the total potential energy.

But then my statement was right. Gauß's law in differential form is not valid for an infinitely stretched constant charge density.

As already stated, this is not what is done when going from the integral form of Gauss’s law to the differential form. You only need the divergence theorem for compact volumes to do that.

But you can perform this transition only when the integrals are well-defined. For the infinitely stretched constant charge density this is not the case. Resp. in other words: In my opinion you can use Gauß's differential form only in those cases, where the integral formulation is also well-defined. The differential formulation is a local formulation which holds within certain domains only, not globally! It holds, where the integral formulation is defined.

 

[Orodruin]

I'm not sure I understand. I thought this was the divergence theorem form compact volumes.

The argument roughly goes as follows: Assume that Gauss's law holds for any compact volume $V$, i.e.,
$$
\frac{Q}{\epsilon_0} = \frac{1}{\epsilon_0} \int_V \rho\, dV = \oint_S \vec E \cdot d\vec S = \int_V \nabla \cdot \vec E dV
$$
where we have only applied the divergence theorem for a compact volume $V$. Now, take a series of smaller and smaller compact volumes $V_i$ containing the point $\vec x_0$, spheres centred at $\vec x_0$ with decreasing radii will do. Essentially by the multi-dimensional equivalent of the mean value theorem for integrals, you find that
$$
\rho(\vec x_i) V_i/\epsilon_0 = (\nabla \cdot \vec E)|_{\vec x = \vec x'_i} V_i
$$
for some points $\vec x_i$ and $\vec x'_i$ in $V_i$. Cancelling the volumes on both sides and noting that both $\vec x_i$ and $\vec x'_i$ must approach $\vec x$ as $i$ increases gives you $\rho(\vec x_0)/\epsilon_0 = (\nabla \cdot \vec E)|_{\vec x = \vec x_0}$. Nowhere is the divergence theorem for the entire space necessary.

Edit: Addition. In essence, the differential version of Gauss's law essentially follows from the infinitesimal version of the divergence theorem, not the extended one.

 

[Metmann]

Nowhere is the divergence theorem for the entire space necessary.

As stated above, the differential formulation holds only locally! It is valid within domains given by the integral formulation, in general not outside of them!

Edit: Addition. In essence, the differential version of Gauss's law essentially follows from the infinitesimal version of the divergence theorem, not the extended one.

What's the infinitesimal version of the divergence theorem? Divergence theorem = Stoke's Theorem = Integrals, at least to my knowledge. Maybe I am missing something.

 

[TeethWhitener]

Nowhere is the divergence theorem for the entire space necessary.

I was saying that the divergence theorem doesn't apply over the whole space. Are you saying it can be done?

 

[Orodruin]

As stated above, the differential formulation holds only locally! It is valid within domains given by the integral formulation, in general not outside of them!

I do not see where you think that I have claimed anything else. Nowhere have I used the global properties. The point $\vec x_0$ in #73 is arbitrary.

 

[Ken G]

Yet we should not lose sight of the fact that Gauss' law in an infinite homogeneous charge distribution with fixed charge-to-mass ratio allows you to pick any origin you like, and correctly deduce the differential equation for the time dependence of the vector displacement r between any two points, which will be d²r/dt² = 4*pi*rho*r/3 for unit charge-to-mass ratio and charge density rho. That is true if you keep all speeds much less than c, and assume an arbitrarily large charge distribution-- it need not be infinite, merely so large that any boundaries are far away compared to the r values of interest. So it would seem that the concern about the mathematics of infinity cannot be part of the physics problem-- physics problems should give reasonable results even when we are agnostic about the possibility of an infinite distribution of charge. So Gauss' law is doing something right, even if the observer cannot know where the "true center" of the charge distribution is, or even if there is any meaning to such a true center (as there isn't when dealing with gravity in a homogeneous universe).

The physics question that I see as remaining unsolved is what will happen to clumps whose charge-to-mass ratio deviates from unity in a truly infinite charge distribution. These clumps (or test charges) could be used to determine the "true field" in a way not possible with gravity, and perhaps one would need to know the boundary conditions to understand what happens to such clumps. One could then say that simply saying you have an infinite homogeneous charge distribution leaves something unspecified, but in a way that only appears if you allow test charges to deviate from the prevailing charge-to-mass ratio. It thus seems that unlike with gravity, here Gauss' law is only sufficient to give r(t) in the homogeneous dynamics, but further information is needed to know what instruments that are equipped with test charges of large charge-to-mass ratio will detect for the "actual field."

 

[Orodruin]

What's the infinitesimal version of the divergence theorem? Divergence theorem = Stoke's Theorem = Integrals, at least to my knowledge. Maybe I am missing something.

By this I mean what follows when you take region around a given point and let its volume approach zero and how this relates to the surface integral. In essence, for a small enough volume $V$
$$
\oint_S \vec F \cdot d\vec S = (\nabla \cdot \vec F) V,
$$
where $S$ is the bounding surface of $V$.

 

[Metmann]

I do not see where you think that I have claimed anything else. Nowhere have I used the global properties. The point ⃗x0\vec x_0 in #73 is arbitrary.

Ah yes, I get your point, sorry, I messed something up. I've got to sort myself.

By this I mean what follows

Ok, didn't know it under this name.

 

[Fishytwofeathers]

So, this has been bothering me for a few days and I'm having trouble understanding where the fault is. If we consider a uniform charge density $\rho$ extending through all space, then by symmetry, I would argue that $\mathbf{E}=0$ in all space. However, this does not agree with what a naive application of Gauss's Law would predict since $\nabla\cdot\mathbf{E}=0\ne\rho/\epsilon$. So where exactly is the argument breaking down? Is there something unusual about describing a vanishing divergence over infinite space?

MMMM, I see your quandary? Perhaps it's the lens from which you are seeing, don't forget you are in the here and now looking back in retrospect? Or perhaps you are not allowing yourself to see it through the lens of Gauss? Perhaps he was the seed, why else would you be asking the question?

 

[NFuller]

Assume that Gauss's law holds for any compact volume VVV,

Isn't the validity of this assumption part of the argument though? I will admit that I am not familiar with some of the mathematical terms you and others are discussing so forgive me if I'm overlooking the obvious.

The origin is arbitrary.

Is the field not unique then?

Newton assumed the universe should be infinite and homogeneous, and he came to the same resolution as NFuller,

Well, if my initial assumption was wrong, I guess it's comforting that I did no worse than Newton.

 

[Orodruin]

Isn't the validity of this assumption part of the argument though?

The question that post answered was ”can you go from Gauss’s law on integral form to the differential form only from its validity for compact volumes?”

Is the field not unique then?

Given an origin and a corresponding boundary behaviour the field is unique. However, if you change the origin you will not have the same boundary behaviour. This is why the boundary behaviour breaks translational invariance.

Well, if my initial assumption was wrong, I guess it's comforting that I did no worse than Newton.

Anyone who has an advanced physics education and says this problem has never puzzled or bothered them is lying or did not think about it enough.

 

[Demystifier]

Even for $\rho=0$, the equation $\nabla {\bf E}=0$ has an infinite number of nontrivial solutions
$${\bf E}=E_x{\bf e}_x+E_y{\bf e}_y+E_z{\bf e}_z$$
where $E_x$, $E_y$ and $E_z$ are arbitrary constants.

 

[Orodruin]

Even for $\rho=0$, the equation $\nabla {\bf E}=0$ has an infinite number of nontrivial solutions
$${\bf E}=E_x{\bf e}_x+E_y{\bf e}_y+E_z{\bf e}_z$$
where $E_x$, $E_y$ and $E_z$ are arbitrary constants.

There are many more solutions than that even. As discussed in posts #35 to #38.

 

[NFuller]

Given an origin and a corresponding boundary behaviour the field is unique. However, if you change the origin you will not have the same boundary behaviour. This is why the boundary behaviour breaks translational invariance.

So if you change the origin the boundary conditions change?

Working off of post #58, let's say that the boundary condition is $\lim_{x\rightarrow\infty}[\mathbf{E}(\mathbf{x})+\mathbf{E}(-\mathbf{x})]=E_{0}\hat{r}$. Then $\mathbf{E}=\rho\mathbf{x}/\epsilon_{0}+E_{0}/2\hat{x}$. The thing I still don't get is where is $\mathbf{x}=0$? Metaphorically speaking, if I had a meter that measured the electric field, and I was somewhere in this space, what would the meter read?

 

[Orodruin]

So if you change the origin the boundary conditions change?

Yes. This would be the very essence of the boundary conditions not being translationally invariant.

Working off of post #58, let's say that the boundary condition is $\lim_{x\rightarrow\infty}[\mathbf{E}(\mathbf{x})+\mathbf{E}(-\mathbf{x})]=E_{0}\hat{r}$. Then $\mathbf{E}=\rho\mathbf{x}/\epsilon_{0}+E_{0}/2\hat{x}$. The thing I still don't get is where is $\mathbf{x}=0$? Metaphorically speaking, if I had a meter that measured the electric field, and I was somewhere in this space, what would the meter read?

This is not the appropriate generalisation of the boundary condition (the correct generalisation would involve integrals of spherical harmonics). Your origin does not have to be anywhere in particular. You are free to pick whatever point you want as the origin and whatever compatible boundary conditions that you prefer. However, there are going to be different choices where only some correspond to the same physical situation.

Regardless, the question in itself is rather unphysical and you will not find a physical situation where it can be applied - even as an approximation.

 

[pierce15]

Forgive me for not having read the past 5 pages of debate, but I have a few thoughts.

First, we can stop arguing about whether this is a problem of the differential form of Gauss's law. The derivation of the differential form from the integral form uses the divergence theorem which holds in this case. In fact you can show that the "charge" has to be 0 everywhere from the integral form alone. Take a cube Gaussian pillbox with two faces perpendicular to our constant E, then Gauss's law in integral form says that $E \cdot A - E \cdot A = Q_\text{in_cube} = 0$, the cube was arbitrary so the charge is zero everywhere.

Anyway, it seems to me that the issue we are having is "how is there a nonzero electric field when there is no charge?" I don't see a problem. You picked an aphysical field and got an aphysical result. The next question is "then why do our toy models of infinite plates and line charges with finite charge density not give aphysical results?" The answer is that they are completely aphysical but model certain situations that can actually occur (i.e. are the limit of other situations), such as the electric field along the axis of a large charged circular laminar surface. On the other hand, the situation in question of a uniform electric field throughout space does not model any realistic scenario.

 

[Metmann]

First, we can stop arguing about whether this is a problem of the differential form of Gauss's law. The derivation of the differential form from the integral form uses the divergence theorem which holds in this case.

I agree, I was totally wrong.

Take a cube Gaussian pillbox with two faces perpendicular to our constant E

E doesn't have to be constant.I think, we should discard every solution that leads to infinite (total) field energy and which breaks the symmetry of the problem (yes, spontaneous symmetry [here I just mean: solution doesn't share the symmetry of the problem] breaking is a well-known concept, but maybe if possible we should try without in classical physics). With symmetry I mean here: Isotropy around every point. If we assume these two conditions, the only solution in both cases ($\rho=0$ and $\rho=\text{const}$) is $\vec{E}=0$, and hence automatically $\rho=0$.
What do you think about this?

 

[tom.stoer]

I think, we should discard every solution that leads to infinite field energy and which breaks the symmetry of the problem (yes, spontaneous symmetry breaking is a well-known concept, but maybe if possible we should try without in classical physics). If we assume these two conditions, the only solution in both cases ($\rho=0$ and $\rho=\text{const}$) is $\vec{E}=0$.
What do you think about this?

$E=0$ does not solve $\nabla E = \rho = \text{const} \neq 0$

The symmetry of the problem is not completely specified w/o boundary condition. Given $\rho = \text{const} \neq 0$ there are two choices:
1) a boundary condition compatible with $\rho = \text{const} \neq 0$; then the allowed solutions obey $E \neq 0$
2) a boundary condition compatible with $E = \text{const} = 0$ following your idea; but then $\nabla E = 0 \neq \rho$, so there is no solution

 

[Metmann]

=0E=0 does not solve ∇E=ρ=const≠0

No, but it would show, that the constant has to be zero, because else the both conditions would not be satisfied.

The symmetry of the problem is not completely specified w/o boundary condition

Well, the symmetry given by constant $\rho$, is: isotropy and homogeneity or equivalently: isotropy around every point.

 

[tom.stoer]

No, but it would show, that the constant has to be zero, because else the both conditions would not be satisfied.

It depends on your starting point. I said "Given ..." so I started with non-zero charge density which was the original question; this rules out maximal symmetry. If you start with maximal symmetry then this rules out non-zero charge.

Well, the symmetry given by constant $\rho$, is: isotropy and homogeneity or equivalently: isotropy around every point.

Obviously not - only in terms of the charge, not in terms of the electromagnetic field. So there is some essential ingredient missing to rule out the mathematical solution for non-zero E; could be a boundary condition, could be finite energy, could be maximal symmetry including E (which is trivial b/c any non-zero E breaks this symmetry :-)

 

[Metmann]

Obviously not - only in terms of the charge, not in terms of the electromagnetic field.

That's what I stated as the second assumption. The E-field is the solution of the equation, so I stated that it should follow the symmetry of the given physical situation, which is represented by $\rho$.

so I started with non-zero charge density

Yes, but if you start with arbitrary constant charge density, my discussion would result in: charge density has to be zero. So with both my assumptions, non-zero charge density would lead to a contradiction.

 

[Orodruin]

Well, the symmetry given by constant ρρ\rho, is: isotropy and homogeneity or equivalently: isotropy around every point.

As has been stated several times over in this thread, the symmetry of the charge distribution is not sufficient to conclude that the problem displays symmetry. In particular, the symmetries of the boundary conditions also need to be checked. There is no way around it. In the case of the constant non-zero charge, there is no boundary condition that is compatible with the differential equation and shows the same symmetries as the charge distribution.

Compare to the finite case of a constant charge inside a sphere of finite radius where the upper half of the sphere is held at potential V0 and the lower half is grounded. The charge distribution clearly displays full rotational symmetry, but the boundary condition does not and therefore your solution will not have full rotational symmetry. You simply cannot conclude symmetry based on the differential equation alone.

yes, spontaneous symmetry [here I just mean: solution doesn't share the symmetry of the problem] breaking is a well-known concept, but maybe if possible we should try without in classical physics

This is a bit off-topic, but why do you think spontaneous symmetry breaking does not exist in classical physics (it does). All it requires is a system with degenerate potential minima and a symmetry of the potential that transform these into each other.

Yes, but if you start with arbitrary constant charge density, my discussion would result in: charge density has to be zero. So with both my assumptions, non-zero charge density would lead to a contradiction.

This only means that one of the assumptions cannot be satisfied. In other words, it rules out solutions to the equation that have constsnt charge density and satisfy your requirements. It does not rule out solutions that do not, but it shows that those solutions cannot satisfy those requirements.

 

[Metmann]

I start from the end:

This only means that one of the assumptions cannot be satisfied. In other words, it rules out solutions to the equation that have constsnt charge density and satisfy your requirements. It does not rule out solutions that do not, but it shows that those solutions cannot satisfy those requirements.

That's exactly what I've written. Constant charge density + requirements = contradiction. Solutions with constant charge density are physically ruled out by the requirements (if one assumes the requirements to be viable, that's another question)

This is a bit off-topic, but why do you think spontaneous symmetry breaking does not exist in classical physics (it does). All it requires is a system with degenerate potential minima and a symmetry of the potential that transform these into each other.

Indeed.

Compare to the finite case of a constant charge inside a sphere of finite radius where the upper half of the sphere is held at potential V0 and the lower half is grounded. The charge distribution clearly displays full rotational symmetry, but the boundary condition does not and therefore your solution will not have full rotational symmetry. You simply cannot conclude symmetry based on the differential equation alone.

In this case, in the interior of the sphere these boundary conditions are also not compatible with the charge distribution and the differential equation. Constant potential implies zero field which implies zero charge density according to Gauß' law, or am I mistaken? Furthermore at the equator of the sphere you would have a discontinuity and hence infinite field. Therefore these boundary conditions in fact cannot be imposed. So in my opinion this situation is not compatible to the present problem, where we want to find a solution within the domain of the charge distribution.

But I see the problem. There are just no meaningful boundary conditions in the present case.

 

[tom.stoer]

That's what I stated as the second assumption. The E-field is the solution of the equation, so I stated that it should follow the symmetry of the given physical situation, which is represented by $\rho$.

The symmetry is encoded in $\rho$ and in the boundary conditions for $E$. $\rho$ alone does not specify the symmetry completely.

Look at the 1-dim toy model. They system and its symmetry can be specified by
1) $\partial_x E = \rho$
2) $E(0) = 0$

This breaks translational invariance and results in $E = x\rho$.

What's wrong with that?

 

[Metmann]

This breaks translational invariance and results in E=xρE = x\rho.

Yes, but if we require that the solution should follow the symmetry of $\rho$ alone, than this implies $\rho = 0$ (or if $\rho \neq 0$ is fixed, there wouldn't be a solution fulfilling the requirement. If the requirement makes sense is another question.)

My way of thinking about the current problem is: Suppose the whole universe (assuming it is infinite) would be filled with a constant charge density, as a background, and no a priori boundary condition is imposed from the exterior of the system (in finite cases humans can always impose a boundary condition, but in the infinite case this is more subtle). Furthermore all further charges in the Universe could be treated as test charges and do not modulate the background distribution in a measurable sense. Is this possible? Approaching this, I would assume that the solution should follow the symmetries of the background. Then the only situation in which this would be possible, would be a vanishing background density.

But now I understand where I lacked understanding all the time. I did not think about boundary conditions, that are imposed within the system due to aspects, not taken into account by the equation alone. All the time I was thinking about the following: The background distribution is the only thing we have and we do not modulate it from without or within the system.

 

[Orodruin]

In this case, in the interior of the sphere these boundary conditions are also not compatible with the charge distribution and the differential equation.

I am sorry, but this is just wrong. It is perfectly possible to find a solution with this boundary condition and charge distribution. It is the superposition of a radial field that grows with the radius (from the charge distribution) and the gradient of the potential field that solves Laplace's equation with the inhomogeneous boundary conditions.

Constant potential implies zero field which implies zero charge density according to Gauß' law, or am I mistaken?

I said constant potential on the sphere. Not inside the sphere. The potential inside the sphere is not a step function, it is a superposition of spherical harmonics multiplied by $r^\ell$.

Furthermore at the equator of the sphere you would have a discontinuity and hence infinite field.

No, this is not the case. The potential step is on the boundary of the sphere. Anywhere inside the sphere you will have a finite field. The field will grow towards infinity as you approach it, but there is no problem with that (the point charge field also does this inside the volume). In general, you should consider these configurations as distributions, not as functions. Besides, the step function is just an example so this is completely besides the point. You can put any function on the sphere that is not constant, $V_0 \cos(\theta)$ works perfectly fine as well and will single out the $Y_{10}$ harmonic in the solution (in fact, the contribution from the boundary condition will be a constant field in the $z$-direction).

Yes, but if we require that the solution should follow the symmetry of $\rho$ alone, than this implies $\rho = 0$ (or if $\rho \neq 0$ is fixed, there wouldn't be a solution fulfilling the requirement. If the requirement makes sense is another question.)

This is the point, this is what is inconsistent. You simply cannot conclude that the problem displays a symmetry unless you take the boundary conditions into account.

 

[Metmann]

I am sorry, but this is just wrong. It is perfectly possible to find a solution with this boundary condition and charge distribution. It is the superposition of a radial field that grows with the radius (from the charge distribution) and the gradient of the potential field that solves Laplace's equation with the inhomogeneous boundary conditions.

I am sorry, too. I missunderstood your boundary conditions. You said "constant charge density inside of the sphere" and then "upper half of the sphere held at constant potential", so I was thinking you would mean, that the interior of the upper half would have this boundary condition, but of course you just mean, that the boundary condition is applied to the sphere itself, not the interior. My fault, sorry.

This is the point, this is what is inconsistent. You simply cannot conclude that the problem displays a symmetry unless you take the boundary conditions into account.

Concerning this, see my post above.

 

[tom.stoer]

You simply cannot conclude that the problem displays a symmetry unless you take the boundary conditions into account.

Of course one could follow a different approach by requiring maximal symmetry instead of specifying boundary conditions.

[We do that in general relativity quite frequently. As an example: we do not require asymptotic flatness but homogenity plus isotropy, resulting in FRW cosmologies; the main difference is that in GR we then find solutions with non-zero density, whereas in electrostatics this is not possible; the main difference is the non-linearity of GR]

 

[Metmann]

We do that in general relativity quite frequently

Probably that's also why my approach here was different. It has been a long time since I've last done classical E-dynamics, while cosmology is all around me.

 

[Orodruin]

that the interior of the upper half would have this boundary condition

That would not be much of a boundary condition now would it?

However, there would a priori not be anything necessarily horrendous about if the potential looked like that. The charge distribution leading to it would be the idealisation of two equal and opposite surface charges approaching zero distance between each other. The field would be described by something like $\vec E = E_0 \delta(z) \vec e_z$ (again, considering the field as a distribution rather than a function). Clearly this is different from a constant charge distribution though.

Of course one could follow a different approach by requiring maximal symmetry instead of specifying boundary conditions.

I agree, but, as you pointed out earlier, that was not the original question in this thread. The original question stated that there should be a uniform non-zero charge throughout all of space.

 

[Metmann]

That would not be much of a boundary condition now would it?

Not in the strict sense ^^ But one could for example consider the situation: You have a some charge density in some domain outside of a sphere and as a "boundary condition" you require the potential to be constant inside the whole sphere. Of course this would immidiately imply vanishing density inside of the sphere, but in this sense one could also use the term "boundary condition" for a condition that is applied to a volume instead of a surface, even though the mathematical concept of boundary condition applies of course only to boundaries, i.e. $d-1$-dimensional domains.

However, there would a priori not be anything necessarily horrendous about if the potential looked like that [...]. Clearly this is different from a constant charge distribution though.

I don't have a problem with the potential ;) It just wouldn't be consistent with a constant charge density, that's the point, just as you say.

The original question stated that there should be a uniform non-zero charge throughout all of space.

Well, even a uniform non-zero charge density is a priori consistent with maximal symmetry (before you solve Gauß's equation). Boundary conditions are something you impose from the exterior. If there exists nothing else in the universe (infinite) than the background charge and small enough test charges, than there is no physical boundary condition and only maximal symmetry restricts solutions to the problems. I think this approach is still an answer to the original question. And within this approach the answer is, that there is no solution with non-vanishing density.

 

[tom.stoer]

I agree, but, as you pointed out earlier, that was not the original question in this thread. The original question stated that there should be a uniform non-zero charge throughout all of space.

I fully agree.

Starting with uniform non-zero charge $\rho$ yields a class of non-trivial solutions for $E \neq 0$.
Starting with maximum symmetry = homogeneity and isotropy yields $\rho = 0$ and $E = 0$.

 

[Orodruin]

If there exists nothing else in the universe (infinite) than the background charge and small enough test charges, than there is no physical boundary condition and only maximal symmetry restricts solutions to the problems.

I disagree with this. A behaviour at infinity is a boundary condition even if it is not posed on a physical boundary. The underlying assumption is that one is making when ignoring infinity is that the contribution from ever more distant shells vanishes. As we have already noted, this is not the case and the potential formally diverges. The boundary condition you put is equivalent to a prescription on how to renormalise the potential. One such way is to introduce a radial cutoff in the integral (formally, subtracting all contributions from radii larger than $r_0$). However, this will naturally break your translational invariance.

 

[Metmann]

A behaviour at infinity is a boundary condition even if it is not posed on a physical boundary.

Sure, no doubt. But in the situation I describe, there is (in my opinion) no physical way to introduce such a condition without yielding $\rho=0$. I mean a meaningful condition would definitely be, that the total energy should be finite (what you do with the cutoff), hence the potential should vanish at infinity. Can you solve $\Delta \Phi = \rho = \text{const} \neq 0$ with $\lim_{|\vec{x}|\rightarrow \infty} \Phi(\vec{x}) = 0$ without a cutoff? Spontaneously I cannot think of any other meaningful boundary condition within this setting.

The boundary condition you put is equivalent to a prescription on how to renormalise the potential. One such way is to introduce a radial cutoff in the integral (formally, subtracting all contributions from radii larger than r0r_0).

But where does this cutoff come frome? In the situation I described, there is no natural cutoff. Of course you can randomly introduce it, but why should that be physical if the charge density is filling the whole universe? And your results depend on the chosen cutoff so its highly subjective. Of course, if you also consider special relativity, you can use as a cutoff the maximal distance a photon could have traveled since BB, but in classical physics? (Yes I know, ED incorporates Lorentz-invariance and hence SR, but let us for know consider we wouldn't know about it. Hmm, but maybe that's also the problem. One probably shouldn't think about electrostatics in purely classical ways.).