# Charge Density at altitude

• brunie
In summary, to calculate the average volume charge density in the layer of air between 500 and 600 m, Gauss' theorem can be used on the thin slab of air between these two altitudes. By assuming the slab to be flat, the electric field can be solved for by using the formula E = kQ / r^2. The resulting charge density can then be divided by the thickness of the slab (100 m) to get the average volume charge density in C/m^3. By integrating the normal component of the electric field over the surface of the volume, the charge contained can also be determined.

#### brunie

In the air over a particular region, at an altitude of 500 m above the ground, the electric field is 150 N/C directed downward. At 600 m above the ground, the electric field is 100 N/C downward. What is the average volume charge density in the layer of air between these two elevations?

charge denisty should be in C / m^3
so given field strength at different altitudes
use
E = kQ / r^2 to solve for Q for each altitude
then
charge density = Q / (4/3π r^3)

radius of Earth is 6378100 m
so respecitive heights are 6378600m and 6378700m

then average these or subtract or sumthing

is this process somewhat correct?

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Use Gauss' theorem on the slab of air between 500 and 600 m. The slab is thin enough that you really don't need to worry about the curvature of the earth. Just pretend it's a flat slab.

Dick said:
Use Gauss' theorem on the slab of air between 500 and 600 m. The slab is thin enough that you really don't need to worry about the curvature of the earth. Just pretend it's a flat slab.

i don't understnd why we can assume it is flat because the question indicates volume, not area

but treating it as a flat slab, then

E = ∂ / 2Eo
where solving for ∂ will give units C/m^2
then it would seem to be logical to divide by 100m (the difference) to acquire C/m^3

but if this is correct to say, which value is used for E?

....

Sorry. Guess I dozed off. By slab I mean some area by 100 m thick. If you integrate the normal component of the E field over the surface of that volume how is that related to the charge contained?

sry, I am not too sure what u mean by integrating over the surface

Did you do Gauss' theorem?