How Many Electrons Transfer in a Capacitor Charging at 460V?

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SUMMARY

The discussion focuses on calculating the number of electrons transferred in a capacitor charging at a potential difference of 460V with a capacitance of 2.56 x 10-8 F. Using the formula q = CV, the charge (q) is calculated as 1.1776 x 10-5 C. The number of electrons is then determined using the elementary charge (e = 1.6 x 10-19 C), resulting in a total of 7.36 x 1013 electrons transferred.

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  • Understanding of capacitor charging principles
  • Familiarity with the formula q = CV
  • Knowledge of elementary charge (e = 1.6 x 10-19 C)
  • Basic algebra for manipulating equations
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This discussion is beneficial for physics students, electrical engineering students, and anyone interested in understanding the fundamentals of capacitors and charge transfer in electrical circuits.

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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.
 
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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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