How Is Charge Density Calculated Between Two Altitudes?

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Homework Help Overview

The discussion revolves around calculating the average volume charge density in the air between two altitudes, specifically between 500 m and 600 m above the ground, using electric field measurements at these heights.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore using electric field values to determine charge density, with references to formulas involving electric field strength and charge. Some suggest applying Gauss' theorem to simplify the problem, while others question the assumptions made about the geometry of the air layer.

Discussion Status

The discussion is ongoing, with various approaches being considered. Some participants have offered guidance on using Gauss' theorem, while others are seeking clarification on the assumptions and methods being discussed. There is no explicit consensus yet on the best approach to take.

Contextual Notes

Participants are navigating the implications of treating the air layer as a flat slab versus considering the curvature of the Earth, as well as the appropriate use of electric field values in their calculations.

brunie
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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?
 

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