How Does Gauss's Law Help Calculate Earth's Total Electric Charge?

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SUMMARY

This discussion focuses on using Gauss's Law to calculate the total electric charge of the Earth based on the electric field in its atmosphere, given as E = 1.50 x 10^2 N/C. The relevant equations include E = kQ/r^2 for point charges and E = kQr/a^3 for spherical charge distributions. The Earth’s radius is specified as 6371 km, and Coulomb's constant (ke) is 8.99 x 10^9 N·m²/C². The confusion regarding the variables 'r' and 'a' is clarified, establishing that 'a' refers to the Earth's radius while 'r' is the distance from the center of the sphere.

PREREQUISITES
  • Understanding of Gauss's Law and its applications
  • Familiarity with electric fields and Coulomb's Law
  • Knowledge of spherical charge distributions
  • Basic algebra for rearranging equations
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  • Study the derivation and applications of Gauss's Law in electrostatics
  • Learn how to calculate electric fields for different charge distributions
  • Explore the implications of electric fields in atmospheric physics
  • Investigate the relationship between electric charge and gravitational effects on Earth
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Students in physics, particularly those studying electromagnetism, educators teaching electric field concepts, and researchers interested in atmospheric electricity and its implications on Earth’s charge.

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Homework Statement


Suppose that the electric field in the Earth's atmosphere is E = 1.50 x 10^2 N/C, pointing downward. Determine the electric charge in the Earth. (The radius of the Earth is 6371 km, and the Coulomb's constant, ke, is 8.99 x 10^9 N · m2/C2.)


Homework Equations


E=kQr/a^3
Q = E(a^3)/kr


The Attempt at a Solution


I have the equation to solve this (E = kQr/a^3) and have rearranged it to Q = E(a^3)/kr, but my problem is that I don't know what r (or a) is. The Earth's radius would be a, right? Or would it be r? If it's r, then what's a, and vice-versa? Please help me figure r and a's values out.
 
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Lolligirl said:

Homework Statement


Suppose that the electric field in the Earth's atmosphere is E = 1.50 x 10^2 N/C, pointing downward. Determine the electric charge in the Earth. (The radius of the Earth is 6371 km, and the Coulomb's constant, ke, is 8.99 x 10^9 N · m2/C2.)

Homework Equations


E=kQr/a^3
Q = E(a^3)/kr

The Attempt at a Solution


I have the equation to solve this (E = kQr/a^3) and have rearranged it to Q = E(a^3)/kr, but my problem is that I don't know what r (or a) is. The Earth's radius would be a, right? Or would it be r? If it's r, then what's a, and vice-versa? Please help me figure r and a's values out.
Judging by the title you chose for this thread, you should be using Gauss's Law to solve this.

Can you state Gauss's Law?
 
E=kQ/r^2. This is under the applications section of the chapter on Gauss's law, but directly after an example problem that says to find the magnitude of the electric field at a point inside the sphere, we use E=kQr/a^3. Is that different from finding the magnitude of the electric field due to a point charge in the center of the sphere?
 
Lolligirl said:
E=kQ/r^2. This is under the applications section of the chapter on Gauss's law, but directly after an example problem that says to find the magnitude of the electric field at a point inside the sphere, we use E=kQr/a^3. Is that different from finding the magnitude of the electric field due to a point charge in the center of the sphere?
The question says nothing about fields inside charged spheres. It mentions the field in the atmosphere (just above ground level, presumably) and charge within the sphere of the Earth.
 
Lolligirl said:
E=kQ/r^2. This is under the applications section of the chapter on Gauss's law, but directly after an example problem that says to find the magnitude of the electric field at a point inside the sphere, we use E=kQr/a^3. Is that different from finding the magnitude of the electric field due to a point charge in the center of the sphere?
E=kQ/r2 is Coulombs Law for the electric field due to a point charge.

E=kQr/a^3 gives the electric field inside a sphere of radius, a, at a distance, r, from the sphere's center, if that sphere has total charge Q which is uniformly distributed throughout the sphere's volume. This has nothing to do with the problem stated here.

Gauss's Law can be used to show that electric field due to a spherically symmetric charge distribution is the same as the field produced by a point charge of the same value as that of the entire sphere.
 

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