Neumann's problem, electrostatics, proof

[fluidistic]

Homework Statement

In the Neumann's problem $\begin{cases} \triangle \Phi =-4\pi \rho \\ \frac{\partial \Phi }{\partial n}=g \end{cases}$, what condition must g satisfy?

Homework Equations

I'm given the answer, it's $\int _ \Omega \rho dV=-\int _{\partial \Omega } g dS$.

The Attempt at a Solution

I'm trying to demonstrate that the given answer is indeed true.
So let Omega be the region of interest (in $\mathbb{R} ^3$, mathematically) and $\partial \Omega$ be its contour.
Now $\int _\Omega \rho dV=Q$, the total charge enclosed in Omega. So I'm left to show that $\int _{\partial \Omega } g dS$ is worth minus the total charge enclosed in Omega. To me it just looks like the Gauss theorem. If $g=\frac{\partial \Phi }{\partial n }=|\vec E _{\text{normal to the surface}}|$, then I must show that $\int _{\partial \Omega } g dS=-Q$.
So I replace g by its value, it gives me $\int _{\partial \Omega } |\vec E _{\text{normal}}| dS=-Q$.
If I use Gauss theorem, I get that $\int _\Omega \rho dV= \int _{\partial \Omega } \vec E \cdot \hat n dS$. (where I used one of Maxwell's equation, $\nabla \cdot \vec E= \rho$, I omitted a constant)
So in order to get what I should, $|\vec E _{\text{normal }}|$ must equal $\vec E \cdot \hat n$. If the normal points inward Omega then I get what I should. However if the normal points outward Omega (as it generally does), then I get that $\int _{\partial \Omega } gdS=Q$ which has a wrong sign.

Do you see any mistake/error(s) in what I've done?

 

[TSny]

I believe your missing negative sign comes from neglecting the negative sign in $\vec{E} = -\vec{\nabla}\Phi$. Thus, $\partial{\Phi}/\partial{n} = -E_n$.

 

[fluidistic]

I see, thank you very much TSny!

 

[fluidistic]

I realize that the problem was either "what does g must satisfy for the solution to exist?" or "what does g must satisfy for the solution to be unique?".
And the answer given is $\int _ \Omega \rho dV=-\int _{\partial \Omega } g dS$ that I showed to be true, but I don't see why this imply that it guaranties that Phi (the potential) either exist (and satisfies Neumann's problem) or is unique (and satisfies Neumann's problem).
Edit: I'm guessing it has to do with if g does not satisfy the relation given as answer, then Phi does not satisfies Poisson's equation.
In this case the original question was probably "what does g must satisfy for the solution to exist?", though I'm not 100% sure.

 

[paranormal]

Your approach is correct, but there is a small mistake in your use of Gauss's theorem. The correct expression is:

$$\int_\Omega \rho dV = \int_{\partial \Omega} \vec{E} \cdot \hat{n} dS$$

Note that the normal vector is pointing outward from the region of interest, so the dot product should be negative. Therefore, the correct expression for Neumann's problem is:

$$\int_{\partial \Omega} g dS = -\int_\Omega \rho dV$$

This matches the given answer and shows that g must satisfy the condition:

$$\int_{\partial \Omega} g dS = -\int_\Omega \rho dV$$

which is equivalent to:

$$\int_{\partial \Omega} g dS + \int_\Omega \rho dV = 0$$

This condition ensures that the total charge enclosed in the region of interest is equal to the total flux of the electric field through its surface.