Posted by Tom on .
In a particular region of Earth's atmosphere, the electric field above Earth's surface has been measured to be 151 N/C downward at an altitude of 280 m and 167 N/C downward at an altitude of 430 m. Calculate the volume charge density of the atmosphere, assuming it to be uniform between 280 and 430 m. (Hint: You may neglect the curvature of Earth. Why?)
I tried using E=pz/ε, where E was 151+167 and z was (430280)/2, half the distance between the ends, and got, 3.75E11 C/m^3, but that was wrong as the online software told me. What should I do?

PHYSICS 
drwls,
Treat the atmosphere as planar and use Gauss' Law. The net flux from an volume of cross sectional area A and thickness t is:
(167  151) *A = 46 Volt/C * A
That equals the charge inside the volume, divided by ε.
46 V/C*A = (charge density)*A*(thickness)/ε
charge density = ε * 46 /t
You should have subtracted the two field strengths, not added them, since E enters the volume on one side and leaves ion the other. The flux is in opposite directions. 
PHYSICS 
Granger,
^^^^drwls
How did you get 167  151 = 46???