Gauss's law: Why does q=0 if E is uniform?

AI Thread Summary
In a region with a uniform positive volume charge density, the electric field cannot be uniform due to Gauss's law, which states that a uniform electric field implies zero divergence and thus zero charge density. If there is a "bubble" within this region where the charge density is zero, the electric field can still be uniform within that bubble. The discussion emphasizes that if the electric field is uniform, it results in zero net flux through a closed surface, indicating no enclosed charge. Therefore, the presence of any charge would disrupt the uniformity of the electric field. Understanding these principles is essential for applying Gauss's law correctly.
Lola Luck
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Homework Statement


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a) In a certain region of space, the volume charge density p has a uniform positive value. Can E be uniform in this region? Explain.
b) Suppose that in this region of uniform positive p there is a "bubble" within which p=0. Can E be uniform within this bubble? Explain.

Homework Equations



E = electric field

Gauss's law: Flux= ∫ E dA = Q/ε0

The Attempt at a Solution



I thought that if the volume charge density p were uniform, E would also be uniform because the charge enclosed by a Gaussian surface would be the same everywhere. However my book says that in a region where "the electric field E is uniform... the volume charge density p must be 0."
 
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Do you know Gauss law in the differential form? That is, with the divergence theorem?
 
If yes: This says that ##\nabla \cdot E = \rho / \epsilon_o## in which case, a uniform E has no divergence, and therefore zero density.

If no: Uniform fields give no flux. You know as a fact that enclosed charges give flux. If you have flux, then the field isn't uniform.

If you wish for something intuitive tell me and I'll try to make something up.
 
Lola Luck said:

Homework Statement


[/B]
a) In a certain region of space, the volume charge density p has a uniform positive value. Can E be uniform in this region? Explain.
b) Suppose that in this region of uniform positive p there is a "bubble" within which p=0. Can E be uniform within this bubble? Explain.

Homework Equations



E = electric field

Gauss's law: Flux= ∫ E dA = Q/ε0

The Attempt at a Solution



I thought that if the volume charge density p were uniform, E would also be uniform because the charge enclosed by a Gaussian surface would be the same everywhere. However my book says that in a region where "the electric field E is uniform... the volume charge density p must be 0."
For part (a):
Ask yourself a related question. Suppose that in some region of space the electric field, E, is uniform . What is ##\displaystyle \oint \vec{E}\cdot d\vec{A}## in that region?
 
davidbenari said:
If yes: This says that ##\nabla \cdot E = \rho / \epsilon_o## in which case, a uniform E has no divergence, and therefore zero density.

If no: Uniform fields give no flux. You know as a fact that enclosed charges give flux. If you have flux, then the field isn't uniform.

If you wish for something intuitive tell me and I'll try to make something up.

Sorry for the late response.
I guess I didn't realize that if there's flux the field isn't uniform, but it makes sense. Thank you.
 
SammyS said:
For part (a):
Ask yourself a related question. Suppose that in some region of space the electric field, E, is uniform . What is ##\displaystyle \oint \vec{E}\cdot d\vec{A}## in that region?

Sorry for the late response.

If E is constant, that integral would equal the product (E)(area). So there's a positive flux, which implies that that region isn't uniform.
 
Lola Luck said:
Sorry for the late response.

If E is constant, that integral would equal the product (E)(area). So there's a positive flux, which implies that that region isn't uniform.
Yes, but that's a closed surface, so if ## \vec{E} ## is constant (in both magnitude and direction), then the flux inward is equal to the flux outward. That's a net flux of zero out of the surface.
 
SammyS said:
Yes, but that's a closed surface, so if ## \vec{E} ## is constant (in both magnitude and direction), then the flux inward is equal to the flux outward. That's a net flux of zero out of the surface.

I see, so when ## \vec{E} ## is constant there can't be any enclosed charge.
 
Lola Luck said:
I see, so when ## \vec{E} ## is constant there can't be any enclosed charge.
That's correct.
 
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