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What happens to Earth's charge when you remove a volume of soil?

  • Thread starter satchmo05
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  • #1
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Homework Statement


If Earth is considered to be metal sphere (radius = 6371[km]), how much charge Q must be deposited on its surface in order for an arc to be established in the air? If surface was charged to this value by removing all electrons from a volume of soil, how large would this volume be? Assume electron density of soil = 7e23[cm-3]


Homework Equations


Coulomb's Law is the only thing that I can imagine working at the moment.
E = (1/(4∏єo)*(q/r2)*ar


The Attempt at a Solution


If I plug in what I know in this equation, I still have the unknown electric field intensity vector and an unknown charge, which is what I'm solving for in the first part of this problem. I do not know how to implement any equation to take into account the "arc established in the air."

If someone could get me going in the right direction, perhaps the correct formula if Coulomb's Law does not apply here. Please let me know. Thanks to all help on this post!
 

Answers and Replies

  • #2
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Alright, so I think I got a bit farther on this question. What I didn't understand at the beginning was what do I do to create the "arc" to be established. My professor informed me that this arc/lightning is due to dielectric breakdown between the soil and the air. The dielectric strength of air is [3MV/m], and I can use Gauss' Law to solve for a spherical symmetrically charged distribution to solve for Q.

So I know Q from the spherical Gauss formula. The problem I am having now deals with the second part of this problem, where I am to find the volume of the dirt if the charge that was found in the last part was removed. The obvious answer here would be to solve for volume in the Density (D) = mass of electron (m) / volume (V). But the charge has to play a part in this answer, but I do not know how.

I would appreciate any help in regards to my problem. Thanks! Best.
 

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