## Electric flux through a closed surface

[EDITED]
Why is the electric flux through a closed surface zero?

My book based the reasoning from a uniform electric field. I understand that for a uniform electric field, the electric field doesn't diminish in magnitude as we move away from the source, and if we construct a box-shaped closed surface not containing the source and parallel to the field lines, the lines entering on one side must leave on the other side. And since the field is uniform, the electric field going in have the same magnitude with those going out.And the effect just cancels.

My question arises for a non-uniform electric field. Suppose that there is a point charge q somewhere in space. We construct the same box-shaped surface which doesn't contain the charge. But now, the magnitude of the electric field going in is different from that going out since the entry and exit points are at different distances from the point charge, and i think that there is a net flux.

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 The formula for gauss' law gives zero, but i think that it is because, in its deriviation, it is assumed that electric fields outside the surface do not influence the flux, and the electric field for every infinitesimal area only comes from the enclosed charge. It would be helpful if the explanation for my problem is not gauss' law. Thanks!! please reply
 Recognitions: Science Advisor The question why things are as they are is usually not answered in natural science, because by definition it's an empirical science. If there is enough empirical evidence it is often possible to make mathematical models about some part of nature, e.g., classical electrodynamics. The fundamental behavior of electric charges and electromagnetic fields is summarized in Maxwell's Equations. One of these basic equations is Gauss's Law, which states that electric flux equals the charge contained in the volume enclosed by the surface chosen (in Heaviside-Lorentz Units): $$\int_{\partial V} \mathrm{d} \vec{S} \cdot \vec{E}=\int_V \mathrm{d}^3 x \rho=Q_V.$$ The surface-normal vectors $\mathrm{d} \vec{S}$ point (by definition!) outwards of the volume $V$ and $\partial V$ is the closed boundary surface of the volume. As an exercise try this for a point charge at rest and a ball with center around it. The field is the Coulomb field (with the charge at the origin) $$\vec{E}=\frac{Q}{4 \pi r^2} \vec{e}_r.$$ You'll see that for any ball, no matter which radius you chose, you'll always get $Q$ as a result of the surface integral. From this you can derive also the result that it doesn't matter that you take a ball. You can take any shape of a volume, as long as the charge is placed inside, you'll get always $Q$ as the result of the surface integral. To prove this, you better translate Gauss's Law first into its local (or differential) form. Since the integral form holds for any volume and boundary surface, by taking an infinitesimal volume, you get $$\vec{\nabla} \cdot \vec{E}=\rho.$$ Then you can use Gauss's integral theorem $$\int_V \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot \vec{E}=\int_{\partial V} \mathrm{d} \vec{S} \cdot \vec{E}.$$ Which is again Gauss's Law in integral form. Now you apply this to an arbitrary volume with the charge taken out by a little ball completely inside the volume. Since in our problem there's no charge except the point charge at the origin, the volume integral gives 0, and the surface integrals give that the integral over $\partial V$ equals in magnitude the integral over the sphere around the charge, which give $Q$. Now you can consider any charge distribution as the sum over many point charges, which proves the theorem for a general charge distribution. That's so simple because the electromagnetic interaction at this very fundamental level is a linear theory, i.e., you can superpose any solutions of Maxwell's equations to get another solution. Thus you get from Coulomb's Law for point charges to the general validity of Gauss's Law.

## Electric flux through a closed surface

I don't really understand your explanation, mainly because its too complicated for me yet.

I think that I need to clarify my question too. I'll edit it and post it again.
The math I'm familiar with, as of now are just algebra,trig and diff and integ calculus. Partial derivatives and vectors weren't yet taught to us. i know a little but it isn't enough