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

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A capacitor with a capacitance of 2.56 x 10^-8 F charges to a potential difference of 460 V, resulting in a charge of 1.1776 x 10^-5 C. The number of electrons transferred during this process is calculated using the formula q/e, where e is the elementary charge (1.6 x 10^-19 C). This calculation yields a total of approximately 7.36 x 10^13 electrons. The discussion highlights the importance of knowing the value of the elementary charge when solving capacitor problems. Understanding these concepts is crucial for accurately determining electron transfer in capacitors.
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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!
 
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