Consider 2 concentric thin hollow conducting shells with charges $q,2q$ and radii $r,2r$ for simplicity. If I attach a conducting wire between the 2 shells, then how would the charges now be redistributed? (what to do can be found in the link below)

Now my question, is why are we not simply adding the total charge and dividing it equally to the 2 objects? I know we have to ensure the charge is redistributed such that the potential is equal but I also remember doing questions like

"if 2 objects having $6 \;\text{C}$ and $12 \;\text{C}$ charges come in contact and pulled apart then what is the charges on each of them?"

The answer I remember being thought is that "the charges they have is $(12+6)/2 \;\text{C}$ each as charge redistribution occurs until their charges equal" why is this not done in the previous case

Here are some similar questions below, which explains what we should do but however doesn't answer why we can't do my 2nd "method".

How to find the distribution of charge on two spheres connected by a conducting wire?

Concentric shells of charge


The moment you connect the shells you no longer have two objects, but one single conductor with a total charge $q + 2q$. So your problem reduces to find the distribution of charge on a charged conductor. The charge will go to the external surface, so as to spread out as far as possible (it is actually done to minimize the energy).

When the shells are not connected, the charge on the inner sphere would love to spread even farther, but it has no means to do so. The connection with the outer shell provides a way.

The difference compared to the case of the two separated objects where one object is not inside the other, is that the 'external surface' of the composite object belongs in its entirety to the outer shell, while when there are two independent objects, the 'external surface' of the composite object is shared between them.

  • $\begingroup$ Thanks for your efforts, but I can't understand the last line. Could you break it into smaller sentences? (Also I suggest using Grammarly, assuming English isn't your first lang just like me. It made things a little easier for me) $\endgroup$ Jun 19 at 15:46
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    $\begingroup$ @AdilMohammed the lack of sleep deteriorates my English. I see the sentence has been edited already, is it still obscure to you? What I meant is that when the objects are side by side, the area of the total external surface is the sum of the areas of the two surfaces; when one object is inside the other, the area of the total external surface is the area of the surface of the outer body alone. $\endgroup$
    – Peltio
    Jun 19 at 17:29
  • $\begingroup$ Yes, i can understand. so if i have a sphere with 100 $m^2$ and a sphere with 50 $m^2$ touch externally (ie the last paragraph case) then the external surface is 150 $m^2$ and also the charge is evenly spread in this 150 $m^2$. I can easily find the surface charge density if I know the total charge on both spheres. So to find the charge on each sphere after separation i can simply multiply their respective areas with this charge density right? $\endgroup$ Jun 20 at 5:20
  • $\begingroup$ I am assuming when my teacher asked about "if 2 objects having 6C and 12C charges come in contact...?" type of question, she probably meant identical objects. in which case the answer $(12+6)/2C$ becomes correct right $\endgroup$ Jun 20 at 5:24
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    $\begingroup$ but when I have two spheres where one is inside the other, I get all the charge going outsides and nothing inside. Thanks! Got it now $\endgroup$ Jun 20 at 5:26

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