Okay maybe a repeated one. I am asking this because none of the other explains how charges redistribute.
Okay here's the problem.
Suppose I have a $4mF$ capacitor and a $2mF$ capacitor. I charge the 4mF capacitor to a charge $Q$, then remove the battery. Now I connect the uncharged $2mF$ capacitor to the charged $4mF$ one. How does the charges on both the capacitor change?
My argument is this:
The charges on the 4mF capacitor will be equally divided between the 2mF and 4mF, that is Q/2 on each of them. Because if we take one plate of each capacitor and connect them, then the charge will be equally divided. That is if we take the +Q charged plate and connect it to one uncharged plate of the 2mF, then each one will have +Q/2 charge. Similarly taking the negatively charged plate we get -Q/2 on each of the plates. Now if we rejoin them we will have two capacitors each charged to Q/2.
But, obviously my argument is wrong and please explain why?
Best Answer
The potential across either capacitor must be equal. The potential difference of the original capacitor is $\frac{Q}{4*10^{-3}}$V Say the 4mF capacitor loses a charge $x$ which is gained by the 2mF capacitor. Then, $$\frac{Q-x}{4*10^{-3}}=\frac{x}{2*10^{-3}}$$
Solving for $x$ we get $x=\frac{Q}{3}$. So the final charge on the 4mF capacitor is $\frac{2Q}{3}$ and on the 2mF capacitor it's $\frac{Q}{3}$.