# Homework Help: Cube in an electric feild

1. Mar 7, 2005

### Gale

In a region of fair weather, the electric feild is found to be vertically down. At an altitude of 400m E=20 N/C at 200m E=90 N/C. Find the net charge contained in a cube 200m on and edge, with horizontal faces at altititudes of 200m and 400m.

I wasn't sure where to start so i used coulumbs law and plugged in the altitudes and electric feilds.

$$20=\frac{kQ}{400^2}$$ and $$90=\frac{kQ}{200^2}$$

Then i found the difference and got .4e6. I thought that would be the answer maybe, but its mult choice and thats not one of them. So now i'm not really sure what to do....

2. Mar 7, 2005

### chroot

Staff Emeritus
Do you know Gauss' Law?

- Warren

3. Mar 7, 2005

### MathStudent

The correct relation to use is Gauss's law. What would be the net flux through such a cube? What is the relation of the net flux through a closed surface, to the charge enclosed (Gauss's Law)?

4. Mar 7, 2005

### dextercioby

Do you know Gauss' law...?You might use it.I'm getting something like 5*10^{17}C...

Daniel.

5. Mar 7, 2005

### Gale

i though gauss's law could only be used when the charge was located inside the surface... oherwise the flux was zero.... but mk... let me look it up again.

6. Mar 7, 2005

### chroot

Staff Emeritus
The charge is inside the surface of the cube. That's what the question's asking for -- how much charge is within the surface of the cube.

And Gauss' law says there's no way to have non-zero flux through a closed surface around zero charge. If you have a non-zero flux through a closed surface, there must be some charge inside it.

- Warren

7. Mar 7, 2005

### Gale

OOoOOhh hmmm.... i didn't realize that at all... no wonder i was so confused. So if the flux=EA and E= kQ/r^2 then.... hmm, what am i solving for... Q? ok then so flux equals 0... umm

$$0=\frac{KQA}{r^2}$$

which means.... umm do i need to integrate somewhere..... i'm getting that Q has to equal zero... i'm really lost i think...

8. Mar 7, 2005

### Gale

ok, so just kidding gauss's law is
$$\Phi=\oint{EdA}=Q/\epsilon_0$$

Soo... I've really got no idea how to use this... i'll stare at my book a while and notes... sorry... i didn't realize i was so lost...

9. Mar 7, 2005

### chroot

Staff Emeritus
Okay. In English, Gauss' law just says that the electric field's "flux" through a closed surface is proportional to the charge inside that surface.

The "flux" of the electric field is the strength of the electric field multiplied by the area over which the field is present. (It's actually a dot product, but when things are perpendicular, as they will be in many problems, it reduces to a simple multiplication.)

So, the first step is to figure out the electric flux through your cube.

The problem says there's an electric field present on the top surface of the cube (so there's some flux there) and on the bottom surface of the cube (so there's some flux there, too). The sides of the cube are perpendicular to the electric field lines, so there is no flux through the sides. All you need to worry about is the top and bottom surfaces of the cube.

The two surfaces each have an area of 40,000 m^2.

The electric field through the top one is 20 N/C into the cube, so there's a flux of 40,000 m^2 * 20 N/C = 800,000 (with some ugly units) coming into the cube.

The electric field through the top one is 90 N/C out of the cube, so there's a flux of 40,000 m^2 * 90 N/C = 3,600,000 (with the same ugly units) coming out of the cube.

There's more flux coming out the cube than going in -- the net flux is 3,600,000 - 800,000 = 2,800,000. Gauss' law says that this flux is proportional to the charge contained inside the cube, like so:

$$\Phi = \frac{Q}{\epsilon_0}$$

So the charge enclosed is...