No Electric Charges if No Electric Field in Region

[putongren]

Homework Statement

If the electric field in a region in space is 0, can you conclude that no electric charges are in that region?

Relevant Equations

Gauss' Law. The net electric flux through any hypothetical closed surface is equal to 1/ε0 times the net electric charge enclosed within that closed surface. The closed surface is also referred to as Gaussian surface.

This is a conceptual question. I think we can conclude that electric charges cannot be present if there is no electric field in that region. Is this an application of Gauss' Law? A net electric flux thru a surface indicates that there is a charge within that region. An electric field must be present within the region if there is an net electrical flux. But since there is no electric field, then a net electric flux cannot exist and thus there is no charge.

 

[kuruman]

I can think of a region in space in which the electric field is zero, yet there are charges nearby. Can you think of such a region?

 

[putongren]

I can think of a region in space in which the electric field is zero, yet there are charges nearby. Can you think of such a

How about a sphere with constant charge density on the surface? The electric field is 0 on the inside of the sphere, but there are charges on the surface.

 

[kuruman]

How about a sphere with constant charge density on the surface? The electric field is 0 on the inside of the sphere, but there are charges on the surface.

You read my thoughts. A closed but arbitrarily shaped conducting surface with charge on it also works. No need for uniform charge density.

 

[putongren]

I'm just wondering, the charges are on the surface, and the electric field is inside the sphere. Does the charges qualify as in the same region as the region inside the sphere? The original question states that the charges has to be in the same region as the electric field.

 

[jbriggs444]

I'm just wondering, the charges are on the surface, and the electric field is inside the sphere. Does the charges qualify as in the same region as the region inside the sphere? The original question states that the charges has to be in the same region as the electric field.

@kuruman shifted the goalposts just a bit. I agree that zero field in the interior of a spherical shell and non-zero charge density on the shell does not qualify as the "same region".

One thing that comes to mind but is definitely non-physical would be a body that spans all of 3 space with a uniform charge density. This has non-zero charge density everywhere but (arguably) no non-zero field anywhere.

 

[kuruman]

I'm just wondering, the charges are on the surface, and the electric field is inside the sphere. Does the charges qualify as in the same region as the region inside the sphere? The original question states that the charges has to be in the same region as the electric field.

The term "region" is not clearly defined and I grappled with this before I posted earlier. Since no operational definition of "region" was specified, I confess to moving the goalposts a bit because nobody said I couldn't. In any case, the surface charges are on the conductor which conductor is arguably the same "region" where the electric field is zero.

 

[PeroK]

I'm just wondering, the charges are on the surface, and the electric field is inside the sphere. Does the charges qualify as in the same region as the region inside the sphere? The original question states that the charges has to be in the same region as the electric field.

If we assume a finite number of point charges, then there must be a non-zero electric field in a small enough neighbourhood of each charge.

 

[kuruman]

If we assume a finite number of point charges, then there must be a non-zero electric field in a small enough neighbourhood of each charge.

Now that is moving the goalposts. The implicit assumption in classical electrostatics is that charge is a continuous fluid, i.e. there is no point-charge "granularity." A charge element $dq=\rho~dV$ abuts the next such charge element.

 

[vanhees71]

I understand the question such that it is assumed that $\vec{E}(\vec{x})=0$ for all $\vec{x} \in V$. From Gauss's equation then you get of course $\rho=0$. It's pretty much a one-liner to answer this question.

Of course the other way, i.e., if $\rho(\vec{x})=0$ for $\vec{x} \in V$, you cannot conclude that $\vec{E}=0$ in this region, which is also pretty obvious, as @PeroK said in #8.

 

[kuruman]

I understand the question such that it is assumed that $\vec{E}(\vec{x})=0$ for all $\vec{x} \in V$. From Gauss's equation then you get of course $\rho=0$. It's pretty much a one-liner to answer this question.

Of course the other way, i.e., if $\rho(\vec{x})=0$ for $\vec{x} \in V$, you cannot conclude that $\vec{E}=0$ in this region, which is also pretty obvious, as @PeroK said in #8.

That's fine, mathematically. The question, as stated, talks about a "region". The issue is whether surface $S$ that encloses $V$ in the case of a charged conductor is part of this region or not.

 

[PeroK]

Now that is moving the goalposts. The implicit assumption in classical electrostatics is that charge is a continuous fluid, i.e. there is no point-charge "granularity." A charge element $dq=\rho~dV$ abuts the next such charge element.

It was more providing a solution for a specific case.

PS to be pedantic, in the continuous case there is no next volume element.

 

[vela]

I understand the question such that it is assumed that $\vec{E}(\vec{x})=0$ for all $\vec{x} \in V$. From Gauss's equation then you get of course $\rho=0$. It's pretty much a one-liner to answer this question.

It's a one liner if you know Gauss's Law as $\nabla\cdot \vec E = \rho$, but I'm guessing the OP is only familiar with the integral form so far. Using the integral form, there's a little more reasoning needed to argue there's no charge in the region.

The issue is whether surface $S$ that encloses $V$ in the case of a charged conductor is part of this region or not.

I'd say it doesn't. The field changes discontinuously at the surface, so the condition $\vec E = 0$ doesn't hold there.