Micro/macro collection between electrostatic

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To determine the fraction of electrons removed from a charged copper ball, one must first calculate the total number of electrons in a neutral 2.00 mm diameter ball using copper's density and atomic properties. The charge of the ball, +70.0 nC, indicates how many electrons have been removed, as each electron has a charge of approximately -1.6 x 10^-19 C. By dividing the number of electrons lost by the total number of electrons in the ball, the fraction can be found. The discussion highlights the need for accurate calculations based on physical properties of copper to solve the problem effectively. Understanding these concepts is crucial for solving electrostatic problems involving charged objects.
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i have question, if A 2.00 mm-diameter copper ball is charged to + 70.0 nC. What fraction of its electrons have been removed?

I know i have to use the relation, q = (Nprotons - Neletrons)e, but i can't figure out the answer.
what does e stand for.
please help
thankyou
 
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e is the charge of a proton.
 
thank
i still need help for the question though
 
It's easy to figure out how many electrons were lost. But if they want to know the fraction of electrons removed, then you'll have to know how much it started with. Using the density of copper and its atomic number and weight from the periodic table, one could figure out how many electrons there would be in a neutral 2mm diameter ball.

I'm not sure if that's the best way to do this problem though.
 

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