# No Electric Charges if No Electric Field in Region

• putongren
In summary: I get what you're saying. The issue is whether surface ##S## that encloses ##V## in the case of a charged conductor is part of this region or...The surface encloses ##V## and therefore is part of the region.The surface encloses ##V## and therefore is part of the 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.

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?

berkeman and hutchphd
kuruman said:
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.

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putongren said:
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.

berkeman
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.

putongren said:
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.

PeroK
putongren said:
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.

SammyS and phinds
putongren said:
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.

jbriggs444
PeroK said:
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.

DaveE and PhDeezNutz
vanhees71 said:
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.

kuruman said:
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.

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vanhees71 said:
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.

kuruman said:
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.

PeroK

## What does it mean if there are no electric charges in a region?

If there are no electric charges in a region, it means that there are no particles or objects within that area that possess a net electric charge. This implies that the region is electrically neutral and does not generate an electric field on its own.

## Can there be an electric field in a region with no electric charges?

Yes, an electric field can exist in a region with no electric charges if it is created by charges located outside that region. The electric field can propagate through space and affect areas that do not contain the source charges.

## How can we detect the presence of an electric field in a charge-free region?

The presence of an electric field in a charge-free region can be detected using a test charge. If a small test charge placed in the region experiences a force, it indicates the presence of an electric field. The direction and magnitude of the force can provide information about the electric field's properties.

## What is the relationship between electric field and electric potential in a charge-free region?

In a charge-free region, the electric field is related to the electric potential by the gradient. Specifically, the electric field is the negative gradient of the electric potential. This means that the electric field points in the direction of the greatest decrease of electric potential.

## Is it possible for the electric field to be zero in a region with no electric charges?

Yes, it is possible for the electric field to be zero in a region with no electric charges. This can occur if the electric field contributions from external charges cancel each other out exactly at that point, resulting in a net electric field of zero.

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