# Gauss's Law Question

1. Oct 4, 2007

### desaila

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.

2. Oct 4, 2007

### learningphysics

Can you post the question word for word? Is this about a sphere? a conducting sphere?

3. Oct 4, 2007

### desaila

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. Oct 4, 2007

### learningphysics

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. Oct 4, 2007

### desaila

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!