Gauss' law: Cause and effect relationship on the flux

[Zayan]

Homework Statement

I am simply confused by the cause and effect relationship on the flux. I know flux of any vector is the dot product of the vector with the area vector. I also know by gauss law, net flux through any surface is the charge enclosed by epsilon naught. Now suppose there's a closed surface and there's a non uniform electric field. So there is a non zero net flux through the surface. So by gauss law the flux equals to charge inside that surface by epsilon naught. Where did that charge come from? Is the charge the cause of the electric field or the effect? What's causing the flux? The field or the charge? I simply don't understand the causality.

Relevant Equations

Flux= q/Eo

It's simply a theoretical question.

 

[haruspex]

there's a closed surface and there's a non uniform electric field. So there is a non zero net flux through the surface.

That doesn't follow. If you place a point charge next to a sphere the field is non-uniform through the sphere but every flux line into the sphere comes out again, so if you integrate over the surface the net flux is zero.

 

[Zayan]

That doesn't follow. If you place a point charge next to a sphere the field is non-uniform through the sphere but every flux line into the sphere comes out again, so if you integrate over the surface the net flux is zero.

I never talked about placing a charge of any sort. I simply said that there's a non uniform electric field in the region which let's say varies with the x coordinate.

 

[haruspex]

I never talked about placing a charge of any sort. I simply said that there's a non uniform electric field in the region which let's say varies with the x coordinate.

The point charge was just an example of a nonuniform field.

 

[Zayan]

The point charge was just an example of a nonuniform field.

Let me rephrase my question. Suppose there's a cube with one corner on the origin. And a non uniform external field in the x direction varying as E=Eo/x. So the field decreases as you move in the x-direction. Hence as the two faces of the cube contributing to the flux are separated by a distance, the field will be different for them. Hence there will be a net flux. And acc. to gauss law, there must be a charge present inside the cube? Does that mean the charge in the cube was the one contributing to the non uniform field or does that mean that the charge is the consequence of the field. Where did the charge come from.

 

[haruspex]

Does that mean the charge in the cube was the one contributing to the non uniform field

Yes.

 

[Orodruin]

No. It means that the charge inside was responsible for creating a divergence in the field. You can have a divergence free non-uniform field. In that case the contained charge is zero.

 

[PeroK]

Let me rephrase my question. Suppose there's a cube with one corner on the origin. And a non uniform external field in the x direction varying as E=Eo/x. So the field decreases as you move in the x-direction. Hence as the two faces of the cube contributing to the flux are separated by a distance, the field will be different for them. Hence there will be a net flux. And acc. to gauss law, there must be a charge present inside the cube? Does that mean the charge in the cube was the one contributing to the non uniform field or does that mean that the charge is the consequence of the field. Where did the charge come from.

Gauss's law is a consequence of Maxwell's equations - in particular that the divergence of the electric field is proportional to the charge density. That implies that any valid electric field has zero flux through any closed surface that does not enclose any charge. This will be true for a non-uniform external field through a cubic or spherical surface.

The field you specified $\vec E = \frac{E_0}{x} \hat x$ has non-zero divergence, so cannot be a vacuum solution to Maxwell's equations.

 

[Orodruin]

In fact, finding the charge distribution given the field is trivial through the differential form of Gauss law: $\nabla\cdot \vec E = \rho/\varepsilon_0$
Just multiply both sides by $\varepsilon_0$

 

[haruspex]

No. It means that the charge inside was responsible for creating a divergence in the field. You can have a divergence free non-uniform field. In that case the contained charge is zero.

It seems @Zayan did not mean merely a nonuniform field. It is a parallel field of nonuniform magnitude. See posts #2 to #5.

 

[Zayan]

It seems @Zayan did not mean merely a nonuniform field. It is a parallel field of nonuniform magnitude. See posts #2 to #5.

Yes exactly. I meant a parallel field not a diverging.

 

[Orodruin]

Yes exactly. I meant a parallel field not a diverging.

If the field is $\vec E = E(x) \hat x$ then the charge density is simply
$$
\rho(x) = \varepsilon_0 \nabla \cdot \vec E = \varepsilon_0 E’(x)
$$
by the differential form of Gauss’ law.

Note that this is a divergent field. The divergence is non-zero.

 

[Zayan]

Yes exactly. I meant a parallel field not a diverging.

If the field is $\vec E = E(x) \hat x$ then the charge density is simply
$$
\rho(x) = \varepsilon_0 \nabla \cdot \vec E = \varepsilon_0 E’(x)
$$
by the differential form of Gauss’ law.

So my question was that is the charge density inside the cube the cause of the field or the effect

 

[PeroK]

So my question was that is the charge density inside the cube the cause of the field or the effect

The usual way to look at classical EM is that the charges determine the fields. And the fields determine how the charges move.

If you specify a field, then you imply a charge distribution. - which determines the field. You can, of course, use a given field to determine where the charges are. Which is what @Orodruin did in this case. There is no sense in which you could create an electric field without charges and that would cause charges to spring into existence.

 

[haruspex]

If the field is $\vec E = E(x) \hat x$ then the charge density is simply
$$
\rho(x) = \varepsilon_0 \nabla \cdot \vec E = \varepsilon_0 E’(x)
$$
by the differential form of Gauss’ law.

Note that this is a divergent field. The divergence is non-zero.

… if $E'\neq 0$.

 

[Orodruin]

So my question was that is the charge density inside the cube the cause of the field or the effect

It is the cause of the divergent part of the field (and as such that part cannot be considered external). You can always add a divergence free field and obtain the same charge distribution. That part represents a field generated by charges external to the volume.

 

[Orodruin]

I would also add that the differential and integral forms of Gauss’ law are completely equivalent, which is easy to verify through the divergence theorem (aka Gauss’ theorem). It is part of the equations of motion that describe the electromagnetic fields (ie, Maxwell’s equations) where charge and current density form the field sources. If you specify an electromagnetic field, then you can always deduce what the corresponding charge and current distributions must be.

 

[haruspex]

Yes exactly. I meant a parallel field not a diverging.

A parallel field (constant direction) will still be divergent if the field strength varies. Think of it as adding new flux lines between the existing ones as you go along.

 

[Zayan]

A parallel field (constant direction) will still be divergent if the field strength varies. Think of it as adding new flux lines between the existing ones as you go along.

Ohh that means the charges inside add up to the field to make it divergent and contribute to the flux?

 

[haruspex]

Ohh that means the charges inside add up to the field to make it divergent and contribute to the flux?

Yes.

 

[kuruman]

I simply don't understand the causality.

In my opinion there is no causality implied in Gauss's law.
In its integral form it's a recipe to perform two separate tasks and compare what you get:
Task A

1. Subdivide a surface $S$ enclosing a volume $V$ into many small elements $dS$.
2. At each element, find the product of the normal component of the local electric field and the area element $~E_n~dS$.
3. Add all such products over the entire surface; you will get a number.

Task B

1. Subdivide volume $V$ enclosed by surface $S$ into many small elements $dV.$
2. Add the charges in each element over the entire volume.
3. Divide the sum by $\epsilon_0;$ you will get another number.

Gauss's law in integral form asserts that the two numbers you obtain after completing the two tasks are the equal. In other other words, you can use the total flux out of any surface to figure out how much total charge is enclosed without looking inside. Conversely, if you look inside and see how much charge is enclosed, you can figure out how flux goes through the surface.

In differential form, Gauss's law asserts that, at any point in space, the source of a diverging electric field from that point is the value of the volume charge density at that point divided by $\epsilon_0.$ In other words, point charges are sources of diverging electric field lines. Enclose that point with a surface, and you have Gauss's law in integral form.

 

[haruspex]

In my opinion there is no causality implied in Gauss's law.

You are right, of course, but in a practical sense, if you want to create a particular field you would generally have to do it by arranging the charges (including arranging by induction).
E.g. for the parallel but divergent field @Zayan describes you could do it with two parallel, broad, nonconducting blocks of some equal thickness, uniformly and oppositely charged. Ignoring edge effects, the field lines are parallel but change in magnitude progressively through each block.

 

[kuruman]

You are right, of course, but in a practical sense, if you want to create a particular field you would generally have to do it by arranging the charges (including arranging by induction).
E.g. for the parallel but divergent field @Zayan describes you could do it with two parallel, broad, nonconducting blocks of some equal thickness, uniformly and oppositely charged. Ignoring edge effects, the field lines are parallel but change in magnitude progressively through each block.

Yes, Maxwell's equations provide the theoretical basis, but the practical sense is another consideraton. When I was in grad school, the seasoned electronics technician in the physics department became a legend in his own time. After a newbie graduate student ill-advisedly expressed to him the opinion that Maxwell's equations are a good starting point for everything E&M, the old timer's response was "Given Maxwell's equations, design a television set."