What exactly happens when two concentric shells, of different radii carrying different charge say $q$ and $Q$, are connected through a conducting wire?

Will the potential of both the shells be the same? Won't all the charges gather around the bigger shell?

  • $\begingroup$ Any persistent potential difference will make the charges flow forever. $\endgroup$ Apr 28 '17 at 16:29
  • $\begingroup$ yes,but since they are connected by a wire,there potential should be same i think? $\endgroup$
    – Megha
    Apr 28 '17 at 16:32
  • $\begingroup$ My bad, the answer to the first question is yes and the second question is no. You need to calculate the potentials of a spherical charged sphere using the formula of capacitance and then equate them to find out the ratio of charges shared between them $\endgroup$ Apr 28 '17 at 16:39
  • $\begingroup$ okay so then the charge on my smaller shell will be become zero,and all charges will gather around the bigger one $\endgroup$
    – Megha
    Apr 28 '17 at 16:44
  • $\begingroup$ I updated my comment. Sorry, I have been out of touch with basic electrodynamics for a long time that's why it took me some time to realise the exact question. Interesting question though. Tell me if you need a formal answer, I'll post it. $\endgroup$ Apr 28 '17 at 16:45

Will the potential of both the shells be the same?

Yes. This is the definition of an ideal conductor.

Won't all the charges gather around the bigger shell?

Yes, they will. Contrary to what I said in the comments, they will gather up on the surface of the bigger sphere. To show this, consider the electric flux through a spherical surface between the two spheres. This is given by:$$\int\vec E\cdot d\vec A=q_{enc}$$where the enclosed charge $q_{enc}$ is the charge on the inner sphere. Now, due to the spherical symmetry of the charge distribution, the electric field has to point radially outwards and have the same value on the sphere. This gives $$|\vec E(r)|=\frac{q_{enc}}{4\pi r^2}.$$ Now, $$\Delta V=\frac{q_{enc}}{4\pi}\int _r ^R\frac{dr}{r^2}=\frac{q_{enc}}{4\pi}(\frac{1}{r}-\frac{1}{R}).$$

Now, since $\Delta V=0$ and $r\neq R$, $q_{enc}=0$. So, all charges end up on the outer shell.


  1. I employed the same principle as in my comments to conclude it: calculate $\Delta V$ for the configuration and then set it to zero. My mistakes were to assume non-applicable formulas of $\Delta V$.
  2. You were right in your intuition.
  3. This process of connecting the two spheres doesn't conserve energy. The field outside the outer sphere is not changed by this action but inside it drops to zero instantly. So, $\int E(x)^2 d^3x$ is not the same before and after the charge rearrangement. Which it should, because, the energy conservation equation in this case is $\frac{dU}{dt}=-\vec E\cdot \vec j$ and according to the configuration, $\vec j$ must have a radially outward component, along the direction of $\vec E$.

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