A charge distribution has uniform density in the x-y directions and varies with z according to: ρ(z) = ρ0e−|z|/t
where ρ0 and t are constants.
(a) Find the potential V (z) and the electric field E(z)
(b) Sketch them clearly showing their behaviors in the regions |z| ≪ t and |z| ≫ t.
(c) In the region |z| ≫ t, the system should behave as if there were a uniform sheet of charge. Find the effective charge density σ.
I am not certain I'm setting up the integral for the field correctly. This is what I've set up as the integrand (I'm doing this on my phone and will do my best to be clear).
1/4pi(epsilon naught)•(rho(z)•(z-z naught) dx dy dz)/(x^2+y^2+(z-z naught)^2)^(3/2).
This is all times z hat, and integrated from
-L to L, after which I will take the limits as L goes to infinity.
I have obviously discarded x and y hats, because the charge is uniform in those directions and the non- z components will cancel, but should contribute something in the z direction. The quantity raised to the three halves is the separation vector times the denominator of the unit vector r (only the z component of which survived in the numerator).
What I'm getting is a x/sqrt(x^2+(yz terms)), and factoring out a x from under the radical and taking x to positive and negative infinity I get 1.
Left over(ignoring constants in front) is rho(z) (z-znaught)dydz/(y^2+(z-znaught)^2)
Integrating over y, I get rho(z)•(z-znaught)•1/(z-znaught)•arctan (1/y) which when evaluated from negative to positive infinity gives me pi.
So now I'm left with an integral of just rho(z) with no terms for distance which seems strange.
The Attempt at a Solution
I guess I put my attempt above. Oops. At any rate, is this a sensible way to go about this? I suppose considering I'm essentially integrating over expanding spheres a term of pi makes sense, but I can't find any similar examples to corroborate my approach. I haven't attempted the second and third parts yet but who knows what that will yield. I'd appreciate any critique or suggestions regarding methodology or set up, whether my approach is yielding what looks like a correct answer or not (it seems somewhat cumbersome but maybe it just is). In the future I will try to actually format this on my phone.