# Question about surface integrals on E&M

1. Feb 2, 2010

### Juan Pablo

I've seen on most books and in class that Gauss's law is usually stated like

$$\oint \vec{E} \cdot d\vec{A} = \frac{q_{en}}{\epsilon_0}$$
Shouldn't the integral be a surface integral rather than a line integral? I've seen times in problem resolution where they evaluate the integral as a normal integral rather than a surface one, is there a formal justification for this?

I have another kinda unrelated question. When you choose a Gaussian surface to calculate an electric field, is it chosen arbitrarily? I mean every surface with the desirable properties will do?

Thanks!

2. Feb 2, 2010

### dacruick

if you choose an arbitrary surface its arbitrary. and yes, an infinite amount of surfaces will have the same flux. and that also is a surface integral. I believe that symbol denotes only that the integral's bounds end and begin at the same point. the dA tells you that its a surface integral.

3. Feb 2, 2010

### Delphi51

That is a surface integral because of the dA (element of surface area, sometimes written dS). The circle on the integral symbol indicates integrating over a closed surface that encloses the charge.

You can choose any enclosing surface that you are able to work out. Usually there is only one such surface that is easy to work out - where the E is known and is perpendicular to the surface everywhere.

4. Feb 2, 2010

### vela

Staff Emeritus
I'm guessing that the area-as-a-vector thing is throwing you off. Your textbook probably explains how that works, but basically, for a small piece of a surface, $$d\vec{A}$$ has a magnitude equal to the area of the piece and a direction perpendicular to the surface.
What usually happens is the dot product is being evaluated so that the integral becomes

$$\oint \vec{E} \cdot d\vec{A} = \oint |\vec{E}||d\vec{A}|\cos\theta = \oint {E\cos\theta} dA$$

where $$E=|\vec{E}|$$ and $$dA=|d\vec{A}|$$, and with a clever choice of the gaussian surface, you end up with something like

$$\oint \vec{E} \cdot d\vec{A} = E \oint dA$$

When you take the surface integral, you're now just summing the area of all these little pieces of the surface, and the result is just the total area of the gaussian surface.

5. Feb 2, 2010

### Juan Pablo

That's pretty much what I was asking. Thanks a lot!. But isn't $$\vec{E} \cdot d\vec{A}$$ an abuse of notation for the flux?

6. Feb 2, 2010

### vela

Staff Emeritus
How so?

7. Feb 2, 2010

### Juan Pablo

Apparently I misremembered. Another question, doesn't the electric field varies with distance? How is it possible to take it outside the integral?

8. Feb 2, 2010

### vela

Staff Emeritus
That's where the choice of the gaussian surface comes into play. You take advantage of the symmetries in a problem to make the calculation easier. For example, if you have a spherically symmetric charge distribution, E will be directed radially, and its magnitude will depend only on r, so if you use a sphere as your surface, the magnitude of E will be constant and $\theta$ will be zero, so you can pull E out of the integral.