How Does Earth's Atmospheric Charge Density Vary with Altitude?

[Punchlinegirl]

In a particular region of the Earth's atmosphere, the electric field above the Earth's surface has been measured to be 155 N/C downward at an altitude of 271 m and 175.5 N/C downward at an altitude of 409 m. The permittivity of free space is 8.85 x 10^-12 C^2/Nm^2. Neglecting the curvature of the earth, calculate the volume charge density of the atmosphere assuming it to be uniform between 271 m and 409 m. Answer in units of C/m^3.

First I figured out the 2 equations for flux and set them equal to each other. \Phi = q_e_n_c / E_o and \Phi = E cos \theta * A Since theta= 90, cos 90=1, so it's just * A.
To find the area of the sphere, I took the area of the larger sphere and subtracted the smaller one.
so A= (E_R * 4\pi R^2)- (E_r * 4\pi r^2)
So then q_e_n_c/ E_o = (E_RA- E_rA)
q_enc= (E_RA- E_rA)*E_o
since it's charge per volume, you divide each side by the volume and solve for the charge.
so q_e_n_c = (E_RA - E_rA)*E_o / 4/3 \pi (R^3-r^3)
i was a little unsure of what each variable was in the problem, but I think
E_r = 155 N/C
E_R = 175.5 N/C
r= 271 m
R= 409
So plugging in gives (155* 4pi (271)^2)=1.43 x 10^8 for E_rA
E_RA = (175.5 * 4pi (409)^2) = 3.69 x 10^8
so 2.26x 10^8/ (4/3)pi (409^3 -271^3)) =1.11 C/m^3
Can someone tell me what I'm doing wrong?

 

[TSny]

We are asked to neglect the curvature of the earth. So, consider a large, flat region of the Earth's surface. Imagine a Gaussian surface in the shape of a rectangular box with the top and bottom surfaces of the box parallel to the ground. Let the top surface of the box be 409 m above the ground and the bottom surface of the box be 271 m above the ground. Let the top and bottom surfaces of the box each have area $A$.

Gauss' law can be used to find an expression for the total charge $Q$ inside the box. The expression will involve the area $A$.

Then we can derive an expression for the charge density $\rho$ and find that the area $A$ does not appear in the expression for $\rho$.