How many electrons are involved in a lightning flash

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Homework Help Overview

The discussion revolves around calculating the number of electrons involved in a lightning flash, given a charge of approximately 35 coulombs. The subject area pertains to electrostatics and charge quantification.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore how to relate the total charge to the number of electrons, with one suggesting the use of a formula related to force, while others redirect the focus to the charge of an electron. There is a question about the correct interpretation of the charge value and its implications for the number of electrons.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about charge and the calculation process. Some guidance has been provided regarding the charge of an electron and how to properly calculate the total number of electrons from the given charge.

Contextual Notes

There is a discussion about the sign of the charge, with emphasis on the fact that lightning is negatively charged, which influences the interpretation of the calculations.

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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
 
[tex]-1.6 x 10^{-19}[/tex] electron per coulombs right? so (-1.6 x 10^-19)*35 = [tex]-56 X 10^{-19}[/tex]
 
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:

[tex]-1.6 \times 10^{-19} \frac{coulombs}{electron}[/tex]

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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