Suppose I can induce a charge Q in a conducting sphere by the traditional induction method, then why is it that the charges would be equally shared if I put this sphere in contact with another identical sphere? (resulting in a charge Q/2 in either sphere)

My understanding is that the extra electrons in the first sphere would neutralize the positive charges in the second sphere, leaving a net negative charge in the second sphere. But why do the net negative charges on either sphere at the end of it have to be equal? (Q/2)

What is the theoretical maximum for the charge that can be transferred to another identical sphere?

Thanks in advance!

  • $\begingroup$ Sorry to bump this, but any ideas? $\endgroup$ Commented Sep 6, 2015 at 9:57

1 Answer 1


The electric field inside a static conductor will always be zero, otherwise the conducted electrons/holes/whatever would move in order to make it so. Your two touching conductive spheres make a single, symmetric conductor. The charge distribution that will result in zero electric field within a symmetric conductor must also be symmetric. Ergo, the net charge on each of the two spheres must be equal.

  • $\begingroup$ Thanks! Also, what would be the theoretical maximum for the charge that can be transferred to another identical sphere, assuming that the first charge had a charge Q? $\endgroup$ Commented Sep 9, 2015 at 0:56
  • $\begingroup$ If you start with charge Q, and end up with that charge equally distributed between the two spheres, how much do you think ends up on each? $\endgroup$ Commented Sep 9, 2015 at 0:58
  • $\begingroup$ Q/2, understandably. $\endgroup$ Commented Sep 9, 2015 at 1:32

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