Solving Electric Field of an Insulating Slab

[little neutrino]

Homework Statement

A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x = d and x = -d. The y- and z- dimensions of the slab are very large compared to d and may be treated as essentially infinite.

Homework Equations

The charge density of the slab is given by ρ(x) = ρ0 * (x/d)^2 , where ρ0 is a positive constant. Using Gauss Law, find the electric field due to the slab (magnitude and direction) at all points in space.

The Attempt at a Solution

Attached is a picture of my attempted solution. I don't know why both my answers are off by a factor of 2, could somebody explain?
The answer given is: |x| > d: (ρ0d)/(3ε) ; |x| < d: (ρ0x^3)/(3εd^2).
Thanks![/B]

 

[haruspex]

Unfortunately the image of your working comes out sideways.
It is not recommended to post images of handwritten working (though yours is more legible than most). Please type algebra into the post, preferably with latex.
Anyway... early on you write $V = \pi r^2 (2x)$, then later integrate x from -d to +d. Seems inconsistent.

 

[little neutrino]

Sorry but I don't understand what you mean... Could you elaborate?
I integrated from -d to +d because if |x| > d, there will only be charges on the insulating material until d.

 

[haruspex]

Sorry but I don't understand what you mean... Could you elaborate?
I integrated from -d to +d because if |x| > d, there will only be charges on the insulating material until d.

Think about what your expression $dV = 2\pi r^2 dx$ means for some x. What does that volume consist of? Where is the 2 coming from? Now consider the dV you get for -x. Is it right to integrate x from -d to +d?

 

[little neutrino]

Oh... Is it correct if I visualize it this way: for each dx, there will be dV of 2 circles of area πr^2, one on the + x-axis and the other on the - x-axis, because the line of symmetry is taken at the origin. So I should integrate from 0 to d or -d to 0, because the other side of the pillbox is already taken care of by the equation dV = 2πr^2 dx ?

*sorry about the presentation I'll learn how to use latex soon!

 

[haruspex]

there will be dV of 2 circles of area πr^2, one on the + x-axis and the other on the - x-axis, because the line of symmetry is taken at the origin. So I should integrate from 0 to d or -d to 0, because the other side of the pillbox is already taken care of by the equation dV = 2πr^2 dx ?

Yes, 0 to d is adequate.

 

[little neutrino]

Ok thanks! :)