What is the excess charge on the surface of the earth?

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The discussion centers on calculating the excess charge on the Earth's surface based on its vertical electric field of 117 N/C. The initial attempt to use the equation E = ke*Q/r^2 was unsuccessful, prompting suggestions to apply the formula for electric flux instead. The correct approach involves using the net electric flux, which is the product of the electric field and the surface area of the Earth, equating it to the excess charge divided by the permittivity constant. The permittivity constant is noted as 8.85 x 10^-12. This exchange highlights the importance of using the appropriate formulas in electrostatics calculations.
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Homework Statement



The Earth has a vertical electric field at the surface, pointing down, that averages 117 N/C. This field is maintained by various atmospheric processes, including lightning. What is the excess charge on the surface of the earth?


Homework Equations



E=ke*Q/r^2 Q=Er^2/ke

The Attempt at a Solution



I attempted to plug in E = 117, r= radius of the earch, and ke= 8.99*10^9 into the equation to find Q, but this answer was wrong. Any ideas, thanks.
 
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Why don't you use the formula for Electric flux? The Net electric flux through surface is basically Electric field * (4pi*(radius of Earth ^ 2)) = excess charge / permittivity constant.
 
Ah..thank you.
 
Last edited:
Yeah, permittivity constant is 8.85*10^-12.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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