Electric flux through the earth

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SUMMARY

The discussion centers on calculating the electric flux through the Earth, given its radius of 6370 km and an average downward-pointing electric field of 140 N/C. The correct approach involves using the equation for electric flux, Φ = E · A, where A is the surface area of a sphere. The participant initially calculated a flux of 7.13x10^16 N*m²/C but misunderstood the directionality of the flux, which should be considered as "out of the Earth." This highlights the importance of understanding the sign of electric flux in relation to the field direction.

PREREQUISITES
  • Understanding of electric flux and its mathematical representation (Φ = E · A)
  • Knowledge of spherical geometry, specifically the surface area calculation (A = 4πr²)
  • Familiarity with electric fields and their units (N/C)
  • Basic principles of integration as it relates to electric flux
NEXT STEPS
  • Study the concept of electric flux in detail, focusing on positive and negative flux scenarios.
  • Learn how to calculate the surface area of a sphere and apply it in electric field problems.
  • Explore integration techniques relevant to electric field calculations.
  • Investigate the relationship between electric flux and charge using Gauss's Law.
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding electric field calculations and their implications in real-world scenarios.

sashab
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Homework Statement



The Earth's radius is 6370 km . There is a downward-pointing electric field in the atmosphere above the Earth's surface, of average magnitude 140. N/C .

a) What is the flux of the electric field out of the Earth?
b) What is the charge on the Earth?


Homework Equations



\Phi=\vec{E}\bulletA

The Attempt at a Solution



I figured that this would be a simple problem of the electric flux through a sphere, so I thought if I used the above equation and by plugging in the area of the sphere I would get the right answer for a), but the computer is saying I'm incorrect. :( I got an answer of 7.13x10^16 N*m^2/C. I know that flux involves integration, but I'm only just starting to learn it so I don't know how to use integration in a problem like this yet. Any help would be really appreciated! Thanks
 
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sashab said:

Homework Statement



The Earth's radius is 6370 km . There is a downward-pointing electric field in the atmosphere above the Earth's surface, of average magnitude 140. N/C .

a) What is the flux of the electric field out of the Earth?
b) What is the charge on the Earth?


Homework Equations



\Phi=\vec{E}\bulletA

The Attempt at a Solution



I figured that this would be a simple problem of the electric flux through a sphere, so I thought if I used the above equation and by plugging in the area of the sphere I would get the right answer for a), but the computer is saying I'm incorrect. :( I got an answer of 7.13x10^16 N*m^2/C. I know that flux involves integration, but I'm only just starting to learn it so I don't know how to use integration in a problem like this yet. Any help would be really appreciated! Thanks

That's a good try. Though you should say how you got it. The thing you might be missing is that flux can be either negative or positive. The question is asking for the flux "out of the Earth".
 
Dick said:
That's a good try. Though you should say how you got it. The thing you might be missing is that flux can be either negative or positive. The question is asking for the flux "out of the Earth".
Ohh okay I see where I went wrong. Thanks for the help!
 

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