Say, you have two different charged spheres having different potentials on their surface. Now you connect two of them by a wire. So, after sometimes, both of them will have the same potential on their surfaces. That means some charge has moved to the sphere which had lesser potential. My question is, do these charges came from the wire or the other sphere which had higher potential?
1 Answer
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The potential at the surface of a charged sphere is given by:
$$ V = k\frac{Q}{r} $$
so the voltage $V$ is proportional to the charge $Q$. Since the potential of the larger sphere decreases that must mean the charge on the larger sphere decreases. The same argument tells us that the charge on the smaller sphere must increase.
So charge flows from the larger sphere to the smaller sphere.
answered Jul 9, 2015 at 8:34
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$\begingroup$ +1: Absolutely correct,sir. But where I stumbled thinking this is this: Why is there potential on the surface of the conductor? Of course there must be electric field, which is imparted by the "charges on the surface of the conductor", right? Now, when it is connected to lower potential conductor, it sends its charges towards the lower potential conductor, right? But why? Who is forcing the charges to go to the lower potential conductor's surface?? The field of the sphere over which they lie or the field of the lower potenial conductor? $\endgroup$– user36790Commented Jul 9, 2015 at 8:45
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$\begingroup$ I think I have got my answer. Since there is lower potential & $E = -\nabla\phi$, electric field must have incited the current which ultimately increased the charge of the concerned sphere to make the spheres equipotential. Just want to know, does it apply to any other (weired & quaint) shaped conductor? Thanks:) $\endgroup$– user36790Commented Jul 9, 2015 at 12:10
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$\begingroup$ Yes, positive charge always flows from high potential to low potential, regardless of the shape of the conductors. $\endgroup$– ragnarCommented Jul 9, 2015 at 17:17