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I recently skimmed through my tutor's physics book, it's not the greatest book out there since it doesn't provide much detail about how the physics works rather its more like a compilation for formulas and stuff.

I was wondering about how the following exposition works,

Say that we have two conducting metal balls, call one A with charge $q_{a}$ and radius $r_{a}$ while the other B with charge $q_{b}$ and radius $r_b$. Suppose we connect A and B with a wire. Then what will happen is that the charge will 'balance out' such that A is now $q'_{a}$ and B $q'_{b}$.

By the conservation of charges I know that $q_{a}+q_{b}=q'_{a} + q'_{b}$. What I can't understand is that why my book says $\frac{q'_{a}}{r_{a}}=\frac{q'_{b}}{r_{b}}$.

I suspect that $V'_{a}=V'_{b}$ so by the fact that $V=\frac{k Q}{r}$ we get the desired formula but I'm not too confident with that. Can someone give an inuitive explanation of this?

EDIT: I read somewhere that after the wire is connected the two ball's potential $V$ are the same. Why is that? For me is just so counter-intuitive since both balls have different charge values $q_{a}$ and $q_{b}$.

A crude illustration

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  • $\begingroup$ Are the balls grounded? $\endgroup$ Commented yesterday
  • $\begingroup$ I don't think so, what will it affect? $\endgroup$ Commented yesterday
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    $\begingroup$ Total charge is conserved if the balls are not connected to anything else, but not if they're grounded. The two cases are different. $\endgroup$ Commented yesterday
  • $\begingroup$ Ah, then in my case the balls are not grounded. $\endgroup$ Commented yesterday
  • $\begingroup$ Then the total charge is indeed conserved. $\endgroup$ Commented yesterday

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When you connect two conducting spheres with a wire, the key idea is that charges move until the potentials equalize, not the charges themselves.

For an isolated conducting sphere of radius r and charge q, the potential is

$V=\frac{kq}{r}.$

So when spheres A and B are connected, equilibrium requires

$V'_a=V'_b\quad \Rightarrow \quad \frac{kq'_a}{r_a}=\frac{kq'_b}{r_b}.$

Canceling k gives the relation your book mentions:

$q'_ar_a=q'_br_b.$

Intuitively, larger spheres can “hold” more charge at the same potential, while smaller spheres need less charge to reach that same potential. The system balances out so both spheres share the same potential, even though their charges end up different.

Just know that connected conductors equalize their potentials, not their charges.

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  • $\begingroup$ Ah, so is the $q'_{a} + q'_{b} = q_{a} + q_{b}$ incorrect? $\endgroup$ Commented yesterday
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    $\begingroup$ $q'_a+q'_b=q_a+q_b$ is always true; that's conservation of charge. $q'_ar_a=q'_br_b$ is the equilibrium condition. $\endgroup$ Commented yesterday

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