Charge in a box

1. Sep 13, 2007

yevi

2 infinite boards placed in space in x =5 and x = -5, charged with unformed positive density $$\sigma$$ , between those boards is a spacial distribution of charge $$\rho$$ given: $$\rho$$(x,y,z)=Ax^2 , A const.

There is no charge outside the boards.

I need to calculate the charge in a box with a side=4, located in (0,0,0) parallel to axis.

I want to approach it using gauss. For this I need to find the flom through the box.

My first question:
Does the fields generated by the boards effect the box or they negate each other and only the spacial field effects?

2. Sep 13, 2007

learningphysics

This just seems to be a straight integration problem... integrate charge density over the volume to get the charge... you don't need the flux...

3. Sep 13, 2007

yevi

Ok, no flux.
What about my qustion, do I skip the fields from boards?

4. Sep 13, 2007

Staff: Mentor

All you need to do is find the total charge within the box. No need to worry about fields from outside (or inside!) the box. Are the boards in the box? If not, their charges don't count.

5. Sep 13, 2007

yevi

The integral is from 0 to 4?

6. Sep 13, 2007

Staff: Mentor

That depends on where the box is located. If it extends from (0,0,0) to (4,4,4), then yes.

7. Sep 13, 2007

yevi

Actually it's center in (0,0,0) so my integral should be from -2 to 2?

8. Sep 13, 2007

Staff: Mentor

That sounds right to me.

9. Sep 13, 2007

yevi

I get this:
$$\int_{-2} ^{2}(64Ax^2)$$

$$\frac{256A}{3}$$

10. Sep 13, 2007

learningphysics

Your integrand is wrong. can you show how you got that integral?

Last edited: Sep 13, 2007
11. Sep 13, 2007

yevi

you mean 64? volume of box.... side=4....

12. Sep 13, 2007

yevi

the integrand is Q.
$$\rho$$(v) = $$\frac{dq}{dv}$$

Last edited: Sep 13, 2007
13. Sep 13, 2007

learningphysics

Take a slice at a particular x... the volume of the slice is 4*4*dx... so what's the charge contained in this slice...

14. Sep 13, 2007

yevi

GOT IT.
Thanks again.