# Electric flux through a closed surface

1. Jun 9, 2012

### iampaul

[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.

Last edited: Jun 9, 2012
2. Jun 9, 2012

### iampaul

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.

3. Jun 9, 2012

### vanhees71

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.

4. Jun 9, 2012

### iampaul

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

Last edited: Jun 9, 2012
5. Jun 9, 2012

6. Jun 9, 2012

### vanhees71

Well, then you have to learn this math first. I don't see any way to understand electromagnetism without a minimum of vector calculus. I don't understand, how you can talk about Gauss's Law without knowledge about volume and surface integrals or the divergence of a vector field. I know that this is the biggest problem for beginning physics or engineering undergrads. I've been puzzled about this very much too in the beginning. A nice book, which helps to get a start, is

Anyway, for now perhaps it would help to answer your question better for your needs now, if you'd tell us, how you formulate Gauss's Law without these mathematical tools.

7. Jun 9, 2012

### iampaul

I understand gauss' law from its formula âˆ«Eda. This integral was described to be a surface integral. I don't understand it formally, but it is similar to the ordinary integral i familiar with in calculus 2. The only difference( i think), is that it is taken across a surface, compared to the integrals in calc 2, which are taken across a line. I'm using the book university physics by young and freedman. It explains gauss' law just by using this formula.

8. Jun 9, 2012

### iampaul

I think i get it already.

9. Jun 10, 2012

### nasu

You can get some qualitative understanding considering your cubic box in the radial field of a point charge. Imagine the box oriented with one face perpendicular to the radius connecting the center of the face to the point charge. So the "bottom" of the box is facing the charge. The field "enters" the box through this face. Due to the divergence of the field, the field "exits" through all the other 5 faces (and not only through the opposite face).
The fact that the field decreases is compensated is compensated by increased area so the input flux is perfectly compensated.
As an alternative, consider a "box" cut from a spherical shell oriented so that the lateral faces are parallel with the field lines.

10. Jun 10, 2012

### iampaul

Thanks Nasu!!

That's what i think happens too. I tried to prove it mathematically but it seemed too complicated, until i saw a projection technique in my book for non-spherical gaussian surfaces. I used a similar procedure and it shows that the inward flux exactly cancels the outward flux.

@ vanhees71
thanks for the book recommendation. I'm about to take up a vector analysis course this semester and im planning to use the book as reference.

11. Jun 10, 2012

### Muphrid

Something else to think about: if you're familiar with methods to solve differential equations, you know that equations of a certain form have homogeneous solutions (when the overall DE is equal to constant zero) and particular solutions (which apply for some arbitrary function).

Solving Maxwell's equations on a surface is the same way. The electric field generated by source within the volume bounded by the surface is the particular solution, and you can add any homogeneous solution (corresponding to electric fields generated by charge outside the volume) to the particular solution and still satisfy Maxwell's equations.