# Flux in an electric field

1. Jun 22, 2008

### drfischer1

hello everybody,
this is a question that has been buffeling me for a while. everything is in this image: http://img398.imageshack.us/img398/5213/58136160lu9.jpg [Broken]

i know 2 things:
1. the total (vector) flux is 0 (no charge inside the cube)
2. the flux through each side is $$\int_{S}*E dA$$
so i've tried
(the flux through shaded) = - (all the rest)
and calculating each side separately but i just don't get it right!
iv'e been integrating since yesterday morning and i have no confidence in my integrals anymore... :-(

could someone explain what i'm missing please?

Thanks!

Last edited by a moderator: May 3, 2017
2. Jun 22, 2008

### Nick89

I think this is just a 'latex issue' but the correct equation should be:
$$\phi_E = \oint_S \vec{E} \cdot \vec{dA}$$
(note in particular the use of the vector dot product)

3. Jun 22, 2008

### Multicol

If I haven't missunderstood something there aren't that much integration involved. Everytime you dot with dA you'll get a scalar and a dobbel integral from 0 to 80 cm

4. Jun 22, 2008

### EngageEngage

yes, there wont actually be an integral to evaluate, if you're tricky (but you can do it with an integral if you want). The first thing you have to do is dot the E field with the vector dA. remember that:

$$\vec{dA} = \frac{\vec{n}}{|\vec{n}|}|\vec{dA}|$$
this normal vector is the one that is normal to the surface that you will be integrating over. So, in totality, you get:
$$\phi_E = \int \int_S \vec{E} \cdot \frac{\vec{n}}{|\vec{n}|}|\vec{dA}|$$

So, what is your normal vector, and what is dA?