Electric Field Inside a Gaussian Surface with Point Charge q

[Jaccobtw]

Homework Statement

Gauss's Law States that if there is no charge enclosed by a gaussian surface, then the electric field must be zero. But how is this the case?

Relevant Equations

E = q/A$\epsilon_o$

If I have a point charge q right outside of a gaussian surface, it makes sense that the flux is zero inside the surface because the electric field going in equals the electric field going out. However, how would the electric field be zero inside? Wouldn't it just take on the electric field of that area in space relative to the point charge q?

 

[Delta2]

Gauss's law states that if there is no net charge enclosed by a gaussian surface then the electric flux is zero (and not that the electric field is zero).

If there is no net charge inside, and hence the flux is zero, then we can deduce that the e-field is zero ONLY IF there is some additional symmetry argument. The situation you describe with the charge outside a gaussian surface is not symmetrical, hence mathematically what happens is that from $$\oint \vec{E}\cdot d\vec{S}=0$$ we can NOT deduce that $$\vec{E}=0$$ because there is no symmetry that would allow us to get the $\vec{E}$ outside of the flux integral, so the following implication $$\oint \vec{E}\cdot d\vec{S}=0\Rightarrow \vec{E}\cdot\oint d\vec{S}=0\Rightarrow \vec{E}=0$$ simply is NOT valid mathematically.

 

[Jaccobtw]

Gauss's law states that if there is no net charge enclosed by a gaussian surface then the electric flux is zero (and not that the electric field is zero).

If there is no net charge inside, and hence the flux is zero, then we can deduce that the e-field is zero ONLY IF there is some additional symmetry argument. The situation you describe with the charge outside a gaussian surface is not symmetrical, hence mathematically what happens is that from $$\iint \vec{E}\cdot d\vec{S}=0$$ we can NOT deduce that $$\vec{E}=0$$ because there is no symmetry that would allow us to get the $\vec{E}$ outside of the flux integral, so the following implication $$\iint \vec{E}\cdot d\vec{S}=0\Rightarrow \vec{E}\cdot\iint d\vec{S}=0\Rightarrow \vec{E}=0$$ simply is NOT valid mathematically.

Why did you use double integrals?

 

[Delta2]

Why did you use double integrals?

Well, the flux integral is a surface integral and so its usually a double integral, but you maybe right, I should have used the $$\oint$$ which means surface integral over a closed surface. I will edit my post shortly.

That is not the main point of this btw, the main point is that there is no symmetry in the situation you describe.

 

[Orodruin]

Gauss's law states that if there is no net charge enclosed by a gaussian surface then the electric flux is zero (and not that the electric field is zero).

If there is no net charge inside, and hence the flux is zero, then we can deduce that the e-field is zero ONLY IF there is some additional symmetry argument. The situation you describe with the charge outside a gaussian surface is not symmetrical, hence mathematically what happens is that from $$\oint \vec{E}\cdot d\vec{S}=0$$ we can NOT deduce that $$\vec{E}=0$$ because there is no symmetry that would allow us to get the $\vec{E}$ outside of the flux integral, so the following implication $$\oint \vec{E}\cdot d\vec{S}=0\Rightarrow \vec{E}\cdot\oint d\vec{S}=0\Rightarrow \vec{E}=0$$ simply is NOT valid mathematically.

This point is way too little emphazised in most treatments and a very common source of misunderstanding among students. The usual takeaway from the typical example of a spherical symmetry is ”oh, EA = enclosed charge/epsilon, very convenient” and the symmetry argument preceeding the statement is often forgotten. It needs to be hammered in.