How Is Charge Density Calculated Between Two Altitudes?

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SUMMARY

The discussion focuses on calculating the average volume charge density between two altitudes, specifically 500 m and 600 m above ground level, where the electric fields are 150 N/C and 100 N/C, respectively. The calculation utilizes Gauss' theorem and the formula E = ∂ / (2ε₀) to derive the surface charge density (∂) in C/m², which is then converted to volume charge density (ρ) in C/m³ by dividing by the height difference of 100 m. The participants confirm that treating the air layer as a flat slab simplifies the calculations, despite initial concerns about the curvature of the Earth.

PREREQUISITES
  • Understanding of electric fields and their units (N/C)
  • Familiarity with Gauss' theorem and its application
  • Knowledge of charge density calculations (C/m³)
  • Basic concepts of integration in physics
NEXT STEPS
  • Study the application of Gauss' theorem in electrostatics
  • Learn about electric field calculations in varying altitudes
  • Explore the relationship between surface charge density and volume charge density
  • Investigate the effects of altitude on electric fields in the atmosphere
USEFUL FOR

Students and professionals in physics, particularly those focusing on electromagnetism, atmospheric science, and electrical engineering, will benefit from this discussion.

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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