# How does charge redistribution occur?

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).
• 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? Jun 20 at 5:20
• 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 Jun 20 at 5:24