A small sphere of radius $r_1$ and charge $q_1$ is enclosed by a spherical shell of radius $r_2$ and charge $q_2$. If $q_1$ is non-zero and the two spheres are connected by a wire, then, charges will flow from the inner sphere to the outer one; no matter what the charge $q_2$ is. Why???

  • $\begingroup$ I guess you know the fact that charges always reside on the surface of a solid pure conductor and not inside it. When you connect inner sphere to outer hollow sphere, the inner sphere becomes a part of the hollow sphere. But it's inside the hollow sphere. So, the net charges would come out on the surface of the hollow sphere. $\endgroup$ – Shubham Sep 16 '15 at 16:38
  • $\begingroup$ thanks , it was so simple , may be i simply didn't put much to understand it. $\endgroup$ – shaistha Sep 17 '15 at 13:36
  • $\begingroup$ when i tried to find an answer for this somewhere , i got the explanation as this, according to gauss's law , E between a sphere and a shell is determined by the charge q1 on a small sphere .Hence , potential between them is independent of charge q2.But the thing which i didn't understand in this explanation is , why E is determined by the charge q1. I TOTALLY UNDERSTAND THE CONCEPT GIVEN BY SHUBHAM BUT DON'T UNDERSTAND THE ABOVE explanation . $\endgroup$ – shaistha Sep 18 '15 at 8:58

Let us consider a shell of radius r2. Take another small shell of radius r1 enter image description here


The answer is the difference between the potentials if the two spheres. When the inner sphere is positively charged, no matter if it's bigger or smaller than the charge that resides on the surface of the outer sphere, as long as it's positive, the potential on the inner sphere will be greater than that of the outer sphere. As you know, charges flow from higher potential to lower potential.

So, if the inner sphere was negatively charged, then the charge would flow in the opposite direction.

  • $\begingroup$ That's assuming charge carriers are positively charged. The correct answer to this question should be independent of the sign of charge carriers. $\endgroup$ – Ali Aug 1 '17 at 9:50

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