Calculating Flux in Box with Charged Boards

[yevi]

2 infinite boards placed in space in x =5 and x = -5, charged with unformed positive density \sigma , between those boards is a spatial 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 spatial field effects?

 

[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...

 

[yevi]

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

 

[Doc Al]

What about my qustion, do I skip the fields from boards?

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.

 

[yevi]

The integral is from 0 to 4?

 

[Doc Al]

The integral is from 0 to 4?

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

 

[yevi]

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

 

[Doc Al]

That sounds right to me.

 

[yevi]

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

But the answer is:
\frac{256A}{3}

 

[learningphysics]

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

But the answer is:
\frac{256A}{3}

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

 

[yevi]

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

 

[yevi]

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

 

[learningphysics]

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

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...

 

[yevi]

GOT IT.
Thanks again.