How many Excess Electrons are in a Typical Lightning Bolt?

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A typical lightning bolt carries about 10.0 coulombs of charge, which translates to a significant number of excess electrons. The charge of a single electron is approximately 1.6 x 10^-19 coulombs. To find the number of excess electrons, one can divide the total charge of the lightning bolt by the charge of a single electron. This calculation confirms that the problem is straightforward, despite initial confusion. Understanding the relationship between charge and the number of electrons is key to solving such problems.
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Homework Statement


A typical lightning bolt has about 10.0 C of charge. How many excess electrons are in a typical lightning bolt?


Homework Equations


There's something in my notes that is confusing me, but I think it could be relevant. It says e = 1.6 x 10-19c


The Attempt at a Solution


Well this question sounds much too easy to be confusing, but we just started this chapter today and I don't have many notes.
 
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Hi chops369! :smile:
chops369 said:
A typical lightning bolt has about 10.0 C of charge. How many excess electrons are in a typical lightning bolt?

There's something in my notes that is confusing me, but I think it could be relevant. It says e = 1.6 x 10-19c

You mean e = 1.6 x 10-19C, where e is the charge of one electron, and C is coulombs …

what's the difficulty? :confused:
 
Well I had figured that originally, but I guess I just didn't think it was that simple.
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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