How Well Do You Understand Gauss's Law?

[roflcopter]

Consider Gauss's law: [tex]
\oint \vec{E}\cdot d\vec{a} =\frac{Q_{enc}}{\epsilon_0}[/tex] . Which of the following is true?

A. E Must be the electric field due to the enclosed charge

B. If q=0 then E=0 everywhere on the Gaussian surface

C. If the charge inside consists of an electric dipole, then the integral is zero

D. E is everywhere parallel to dA along the surface

E. If a charge is placed outside the surface, then it cannot affect E on the surface.
The attempt at a solution

I've ruled out B: Because I can have a point charge outside the Gaussian surface and so E is not zero necessarily at the surface since it will create an E field.

I've ruled out D: Because I can have a cube and E will not always be parallel to the 6 sides. Only case I can think of E being always parallel to dA is for a sphere.

I've ruled out E: Because this is similar to B. The external charge will create an E field of E=kQ/r^2.

So I say the answer is C since q(enclosed) will be zero leaving the integral equal to zero. Or, the answer could be A since isn't that kind of the definition of Gauss's law anyways? Or well, I guess not since we could have a Gaussian surface with no charge in it and a charge outside with E=kQ/r^2. So, the E vector in the integral is not necessarily due to the charge inside the Gaussian surface, right?

I don't think I'm supposed to have multiple answers though...not sure. I'm leaning more towards answer C.

 

[Delphi51]

I don't like A. E is the electric field due to all charges in the universe, not just the ones inside. As you said in your comment on B.

 

[nlightner]

I would say that the correct answer is A. Gauss's law states that the electric field flux through a closed surface is proportional to the enclosed charge. This means that the electric field at any point on the surface is due to the charge enclosed within that surface. Therefore, A is the most accurate and comprehensive answer.

Answer B is not always true because, as you mentioned, an external charge can still create an electric field on the surface.

Answer C is not necessarily true. The integral may not be zero if the charge inside consists of an electric dipole because the electric field from the dipole may not be symmetric with the surface.

Answer D is not true because, as you mentioned, the electric field may not always be parallel to the surface.

Answer E is not true because an external charge can still affect the electric field on the surface.