How does charge redistribute when a charged capacitor is connected to an uncharged capacitor?

Question

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?

Answer 1

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 $\frac{Q}{3}$.

Answer 2

Your answer is wrong in the sense that the charge will not be equally divided between the plates of the two capacitors. And the reason is that the capacitors have different capacitances. In this case you need to apply Kirchoff's voltage rule to get the correct equation. If $Q$ is the total charge on the charged capacitor before discharging starts and $Q_0$ is the charge on the uncharged capacitor, then by Kirchoff's voltage rule, we have $$\frac{Q-Q_0}{C}+\frac{Q_0}{C_0}=0$$

Now solve this.

Hope this helps you.

Answer 3

Great question!

• If it were just 2 metal plates, charges would be equally distributed.
• But in a capacitor arrangement, the NET electric field between the plates (due to both plates) contribute to the voltage across them.(Δ V = − ∫ E ⋅ dl).
• When the two are connected, KVL tells us that the sum of voltages across any closed loop must be zero. Therefore charges must redistribute such that voltage is same.

Thanks!

Answer 4

Look my friend,first of all during charging of capacitor ,CHARGED AND UNCHARGED CAPACITOR ARE NOT IN CONTACT.SO EQUAL CHARGE IS NOT TRANSFERRED.Now process of charging of capacitor.From definition, how did you define a capacitor,two plates separated by an insulating material like air.Now I an going to tell you how actually charging of capacitor takes place.

Take a charged capacitor like positive charged and one uncharged capacitor and don't touch them.now induction happens.The end nearer to positive plate will have negative and other have positive polarity.Now you will know that negative charges will reduce the potential of charged plate and positive positive charges will increase its potential. So if we earth or ground the positive charge on inducted plate.Now there is only negative charges.So it will decrease the potential at charged plate all time when its potential becomes maximum due to positive charges.So all time it charges and discharges.

FOR BETTER UNDERSTANDING,Go to YouTube and see examfear capacitance part 6.Watch this and you will understand what I am saying.

Hope it helps.