# Question about electric fields and operators

1. Oct 27, 2008

### Amok

1. The problem statement, all variables and given/known data

I have three questions concerning the electric field:

1- When calculating an electric flux for a spherical charge distribution my proffessor always writes "4 pi r2 E(r) = flux", where E(r) is the electric field. I don't understand this. I've tried to calculate the flux through a sphere by using the divergence theorem and E(r) = kr-2, but I just don't get the same result. I'm not that good at vector calculus, so maybe I'm just doing mistakes when I integrate. Could anyone show me how to do this integration so I can understand this?

2- I don't get why the definition of the electric potential (with the integral) is equivalent to saying E = -grad e(e is the potential of E). I know it involves the gradient theorem, but I don't really get it, and I can't find a decent demonstration anywhere.

3- More or less the same question. Why can I write Gauss's law either with the integral over a surface or by using the divergence? I know it has to do with the divergence theorem, but once again, all demonstrations I've found go to fast for me too understand them.

2. Relevant equations

This is my first post, and I don't really understand how to write equations properly, so I'm posting links to wiki pages that contain the equations for each question:

1- http://en.wikipedia.org/wiki/Gaussian_surface

2- http://en.wikipedia.org/wiki/Electric_potential

3- http://en.wikipedia.org/wiki/Gauss_Law

And more importantly I've attached a word file with all the equations written down properly.

#### Attached Files:

• ###### questions.doc
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2. Oct 27, 2008

### gabbagabbahey

$\int_{\mathcal{GS}} \vec{E} \cdot \vec{da}=\frac{Q_{enclosed}}{\epsilon_0}$
where $\mathcal{GS}$ is the Gaussian surface. This law holds for any charge distribution and hence any electric field; not just the field of a point charge (kq/r^2)....It is however only useful when the charge distribution (and hence the electric field it produces) possesses certain types of symmetry.
The relation, "4 pi r2 E(r) = flux" applies to situations where the field is spherically symmetric about the origin. In such cases, the electric field is radially outward and is uniform (it has the same value for any theta and phi values). Choosing a concentric sphere of radius r as a gaussian surface is therefor useful; the area element of such a surface is $r^2sin(\theta)d \theta d \phi$ and also points radially outward. The dot product between the electric field and the are element becomes just $|E|r^2sin(\theta)d \theta d \phi$. Clearly, since neither |E| nor r^2 vary over the Gaussian surface, they can come outside of the integral. That leaves the integral over theta and phi which just gives 4*pi.