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So I was reading my physics textbook and it said that "Suppose we have two metal spheres, one highly charged and the other electrically neutral." What does it mean by highly charged? Does that mean the sphere are both protons and electrons in it?

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It shows the picture above that there are only protons in the charged sphere? Also when you place a nail on top of both spheres, which acts as a conductor, do the free electrons from the nail attract the protons from the charged sphere and move those protons to the neutral sphere? Since again in part B after putting the nail on top of both spheres, the previously neutral sphere is charged but with only positive protons?

If someone can clear up my confusion that would be great.

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Positively charged means lack of electrons. Remember that protons are stationary only free electrons can move. So if you have excess electrons its negative and if no of protons= no of electrons its neutral.

So what happens in above case is that one sphere has lack of electrons and the other is neutral. By joining them by a conductor some electrons from neutral sphere jumps to the other sphere.

The main thing to understand is protons never flow. Thus charged almost always means either lack of excess of electrons.

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  • $\begingroup$ OHH... Now that makes sense. I forgot that protons never flow. Thank you so much for this answer! $\endgroup$
    – John
    Sep 22, 2015 at 4:29
  • $\begingroup$ Wait actually how do the electrons in the neutral sphere jump to the other sphere? I thought the electrons int he neutral sphere would repel the free electrons in the nail? So therefore, the electrons can't move from the neutral sphere to the other sphere. Correct? $\endgroup$
    – John
    Sep 23, 2015 at 19:21
  • $\begingroup$ Actually at first the electrons of the nail goes to the positive sphere then the nail becomes positive and then the electrons of neutral sphere goes into the nail. If the dimension of nail is negligible then the above pic holds good. Otherwise there will be charges on nail too. $\endgroup$
    – Ari
    Sep 24, 2015 at 4:35

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