0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Sorry to bump this, but any ideas? $\endgroup$ – Ferreroire Sep 6 '15 at 9:57
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Ferreroire Sep 9 '15 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? $\endgroup$ – Daniel Griscom Sep 9 '15 at 0:58
  • $\begingroup$ Q/2, understandably. $\endgroup$ – Ferreroire Sep 9 '15 at 1:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.