# Two Capacitors Problem

## Homework Statement

Two capacitors C1=4μf, C2= 12μf are connected in parallel across 12V battery. Then they are collected in parallel, with positive side of one capacitor to the negative side of the other. Find the resulting potential difference.

## Homework Equations

Cparallel= C1 + C2
1/Cseries= 1/C1 + 1/C2

## The Attempt at a Solution

Firstly they are connected parallel and equivalent capacitance is 16μf and total charge is 192μC (144 + 48). Then they are connected in a combination which is as good as series hence now charges on each capacitor will be same also net capacitance is 3μf.
3*10^(-6)*V= 192*10^(-6) = 64V (not working!!)

gneill
Mentor

## Homework Statement

Two capacitors C1=4μf, C2= 12μf are connected in parallel across 12V battery. Then they are collected in parallel, with positive side of one capacitor to the negative side of the other. Find the resulting potential difference.

## Homework Equations

Cparallel= C1 + C2
1/Cseries= 1/C1 + 1/C2

## The Attempt at a Solution

Firstly they are connected parallel and equivalent capacitance is 16μf and total charge is 192μC (144 + 48). Then they are connected in a combination which is as good as series hence now charges on each capacitor will be same also net capacitance is 3μf.
3*10^(-6)*V= 192*10^(-6) = 64V (not working!!)

When a capacitor is charged with some charge q, it gets +q on one plate and -q on the other. When you flip one capacitor over before reconnecting them, the negatively charged plate of each capacitor is connected to the positively charged plate of the other. What happens to + and - charges when they meet?

When you have only two components making up the entire circuit you have a choice of considering them to be connected in series or in parallel. For this problem you should consider that the capacitors are in parallel; It makes the solution more direct.

What happens to + and - charges when they meet?

When you have only two components making up the entire circuit you have a choice of considering them to be connected in series or in parallel. For this problem you should consider that the capacitors are in parallel; It makes the solution more direct. wink:
According to me the charges will move along the circuit to attend electrostatic equilibrium. But that is not helping me in getting correct answer. Moreover in parallel +ve plate of one capacitor is connected to +ve plate of other and here its exactly opposite. Help me to rectify my conceptual mistake.

gneill
Mentor
You didn't address the question: "What happens to + and - charges when they meet?"

Whether or not two components are connected in parallel has nothing to do with their charges or polarity; it's a matter of physical connections. Two components are connected in parallel if they share exactly two nodes. Any charges will sort themselves out via Kirchhoff!

Sorry but I tried my best to go for this but i am unable to answer it. I even reread the chapter again. And when + and - meet potential difference is created. But i m still not getting answer.

gneill
Mentor
Sorry but I tried my best to go for this but i am unable to answer it. I even reread the chapter again. And when + and - meet potential difference is created. But i m still not getting answer.

When + and - are separated a potential difference is created. Suppose you have a small conductive object charged with +2Q and you add -2Q to it. What's the net charge?

This means that the net charge on the plates connected must be zero. Which implies that charge on both the capacitors must be equal i.e half of the net charge that was stored in both of them when charged through 12V. Am I right ??

gneill
Mentor
This means that the net charge on the plates connected must be zero. Which implies that charge on both the capacitors must be equal i.e half of the net charge that was stored in both of them when charged through 12V. Am I right ??

No, that's not quite right.

In this case the initial charges on each capacitor are different (what are they?). When the charges on the plates that are interconnected meet, equal amounts of positive and negative charges will meet and cancel, but there will be some charge left over. The remaining charges will spread themselves appropriately across the plates.

Yes, originally capacitors had different charges i.e 144 and 48 (in microC). Now equal charges will cancel out and remaining charges will be 144-48 = 96. Right ??

gneill
Mentor
Yes, originally capacitors had different charges i.e 144 and 48 (in microC). Now equal charges will cancel out and remaining charges will be 144-48 = 96. Right ??

Right. So this remaining 96 μC will distribute itself across the two capacitors in the usual fashion -- note that you'll have just positive charges left on one pair of connected plates, and just negative ones left on the other pair.

You should be in a position to determine the potential across the combined capacitors.

This 96microC will distribute according to capacitance ratio as 72 and 24. One capacitor will have +72 & -24 and other will have +24 & -72. But its not getting me to Potential of 3.5V. How to proceed ??

gneill
Mentor
This 96microC will distribute according to capacitance ratio as 72 and 24. One capacitor will have +72 & -24 and other will have +24 & -72. But its not getting me to Potential of 3.5V. How to proceed ??

No, individual capacitors will have the same magnitude of charge on both plates. When we say that a capacitor has a charge Q, it means that it has a charge +Q on one of its plates and -Q on the other -- they are always exactly balanced. In this case there was +96 μC available to distribute between the two capacitors on one pair of connected plates, and -96 μC on the other connected pair; That's what charge cancellation left over on the connected pairs.

So in other words, one capacitor has a 24 μC charge and the other capacitor has a 72 μC charge.

What's the formula relating capacitor charge to potential difference (voltage) on that capacitor? What's the voltage on each capacitor?

Its V=Q/C. The voltage one each capacitor will be 6V assuming that 72microC is the charge on 12microF capacitor.

gneill
Mentor
Its V=Q/C. The voltage one each capacitor will be 6V assuming that 72microC is the charge on 12microF capacitor.

That looks good!

That looks good!

As both the capacitors have 6V. The potential difference is 0V. But the correct answer is 3.5V which is nowhere near my analysis.

gneill
Mentor
As both the capacitors have 6V. The potential difference is 0V. But the correct answer is 3.5V which is nowhere near my analysis.

No, the potential on both is 6V and they are in parallel which means they have the same voltage. The answer is 6V. If your answer key says 3.5 V, then the answer key is wrong --- probably a result of someone changing the principle values in the question but failing to update the answer key.

You should make yourself a drawing depicting the sequence of events that occur and which charges cancel and what remains where in order convince yourself that both capacitors end up with the same potential difference which is oriented in the same direction.

Thanks.

Thanks.