How many electrons are involved in a lightning flash

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A typical lightning flash delivers about 35 coulombs of negative charge from cloud to ground. To determine the number of electrons involved, the charge of a single electron, approximately -1.6 x 10^-19 coulombs, is used. Dividing 35 coulombs by the charge of one electron reveals the total number of electrons in a lightning flash. The calculation confirms that the flash consists of a substantial number of electrons, emphasizing that the charge must be negative due to the nature of electrons. This discussion clarifies the relationship between charge and the number of electrons in a lightning event.
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A typical lightning flash delivers about 35 C of negative charge from cloud to ground. How many electrons are involved?


Im not extactly sure how to approach this question. it should be an easy one since it's one of the first problems from the book.

Im thinking i might need to use this formula: F = qE

since I am trying to find the force right? please help, i am really confuse
 
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Nope, how much charge does an electron have in coulombs? so if you have 35 coulombs then how many electrons do you have?

nothin to do with forces
 
-1.6 x 10^{-19} electron per coulombs right? so (-1.6 x 10^-19)*35 = -56 X 10^{-19}
 
Reality check: how can a lightning flash contain only a tiny fraction of an electron? :eek:

Think of the electron charge as having units as follows:

-1.6 \times 10^{-19} \frac{coulombs}{electron}

Now, how should you multiply or divide this with 35 coulombs, in order for the result to have units of "electrons"?

Actually, the total charge really should be given as -35 coulombs. Since the lightning flash is made up of electrons, it has to be negatively charged. That takes care of the other problem with your answer: it gives a negative number of electrons!
 
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