Capacitor Electron Transfer

  • #1
jacksonpeeble
Gold Member
118
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Homework Statement


A capacitor has a capacitance of 2.56 10-8 F. In the charging process, electrons are removed from one plate and placed on the other plate. When the potential difference between the plates is 460 V, how many electrons have been transferred?


Homework Equations


q=CV
Number of Electrons=q/e


The Attempt at a Solution


q=2.56*10^-6*460=1.1776*10^-5

What is e? I assume this is a constant, but I don't have my book on me to find what it is exactly.
 

Answers and Replies

  • #2
jacksonpeeble
Gold Member
118
0
Nevermind, e=1.6*10^-19, so the final answer is 7.36*10^13. Sorry for the premature post!
 
Last edited:

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