Gauss's Law and electric field Question

In summary, the question is asking if the net flux is zero when a point charge is placed on an arbitrary surface. The answer is that the net flux is not zero, but the electric field at the surface is still not zero because the point charge will create a field.
  • #1
desaila
25
0
This question is giving me problems.

Consider Gauss's law: (integral) E.dA = qenc/(perm. free space). Which is true?

A. If q = 0, then E = 0 everywhere on the surface.
If there is no enclosed charge, there is no flux.. so no electric field, correct?

B. On the surface, E is everywhere parallel to dA.
This is true, at least for spherical surfaces.

C. A charge outside the surface cannot affect E on the surface.
I don't think this is true. Why wouldn't it affect it?

D. E must be the electric field due to qenc.
Even with an enclosed charge, E inside the surface is still 0.

There is also a none of the above option.

Could multiple answers be correct? I think A and B are both true.
 
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  • #2
Can you post the question word for word? Is this about a sphere? a conducting sphere?
 
  • #3
I may have confused things by adding my own comments to it. I will clean it up some. I hope this helps. There is no specific surface type mentioned in the question.

Consider Gauss's law: (integral) E.dA = qenc/(perm. free space). Which is true?

A. If q = 0, then E = 0 everywhere on the surface.
B. On the surface, E is everywhere parallel to dA.
C. A charge outside the surface cannot affect E on the surface.
D. E must be the electric field due to qenc.
E. None of these.
 
  • #4
For part A... assuming that by q = 0, they mean qenclosed is 0... and assuming this is just an arbitrary surface... then it is true that the net flux is 0. But that doesn't mean that E = 0 everywhere on the surface.

Suppose I just have one single point charge Q. And I take an arbitrary surface such that this charge is outside the surface... the charge enclosed in the surface is 0... the net flux through the surface is 0. but the field at the surface is not zero... because the point charge Q will create a field of kQ/r^2...

There is no reason for B to be true... simply construct an arbitrary surface where the field is not perpendicular to the surface...

No reason for C to be true...

No reason for D to be true...
 
  • #5
I wasn't considering external point charges when thinking about A. I now understand why the other statements are incorrect.

This all makes sense now. Thank you!
 

1. What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric field at a point to the electric charge enclosed by a surface surrounding that point. It states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space.

2. What is the significance of Gauss's Law?

Gauss's Law is significant because it allows us to calculate the electric field at a point without having to consider the contributions from all individual charges. Instead, we can simply use the total charge enclosed by a surface to determine the electric field at that point.

3. How is Gauss's Law related to electric field?

Gauss's Law is directly related to electric field as it is a mathematical expression that relates the two. It allows us to calculate the electric field at a point by considering the charge enclosed by a surface surrounding that point.

4. Is Gauss's Law applicable to all types of surfaces?

Yes, Gauss's Law is applicable to all types of surfaces, including closed, open, and curved surfaces. However, it is important to ensure that the surface chosen is symmetrical to make the calculations easier.

5. Can Gauss's Law be applied to both point charges and continuous charge distributions?

Yes, Gauss's Law can be applied to both point charges and continuous charge distributions. For point charges, the electric field can be calculated at a point using the inverse square law. For continuous charge distributions, the electric field can be calculated by dividing the distribution into smaller elements and using integration.

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