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.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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