2 capacitors reaching equillibrium?

[**Mariam**]

Homework Statement

A capacitor 1 of 3nF is charged with 3 volts. When it is completely charged, we connect it to an uncharged capacitor 2 of 7 nF. Calculate the charge on each capacitor after they reach electric equillibrium.

Homework Equations

Q=CV

The Attempt at a Solution

my confusion is how to set up the problem. Like is it the way you solve a series or parallel capacitor diagram? And is the voltage or charge constant between them?

 

[CWatters]

my confusion is how to set up the problem. Like is it the way you solve a series or parallel capacitor diagram?

Start with a diagram. Two capacitors and a switch.

And is the voltage or charge constant between them?

The voltage on each starts off different so it's not a constant voltage problem.
What do you think about the charge? Can any leave the circuit?

 

[**Mariam**]

Start with a diagram. Two capacitors and a switch.
The voltage on each starts off different so it's not a constant voltage problem.
What do you think about the charge? Can any leave the circuit?

And I don't think the charge leaves the circuit.

So for capacitor 1: Q=3*3= 9C
And capacitor 2: Q=0
Total charge= 9C
And if they reach equillibrium the two charge distributes, but is it equal?why?

 

[jbriggs444]

And I don't think the charge leaves the circuit.

Right. Any current that flows to discharge the one capacitor charges the other one. The total charge is unchanged.

And if they reach equillibrium the two charge distributes, but is it equal?why?

In the equilibrium state, no current is flowing. What does this tell you about the potential difference across the two capacitors?

 

[**Mariam**]

Right. Any current that flows to discharge the one capacitor charges the other one. The total charge is unchanged.

In the equilibrium state, no current is flowing. What does this tell you about the potential difference across the two capacitors?

Is potential difference 0?

 

[jbriggs444]

I am speaking of the potential difference across each capacitor individually. One or both of those capacitors will be carrying a charge. The potential difference across a charged capacitor will not be zero.

 

[**Mariam**]

I am speaking of the potential difference across each capacitor individually. One or both of those capacitors will be carrying a charge. The potential difference across a charged capacitor will not be zero.

Oh ok.

But I still don't get the question

 

[cnh1995]

Oh ok.

But I still don't get the question

When the current is zero, what will be the net voltage in the circuit?

 

[**Mariam**]

When the current is zero, what will be the net voltage in the circuit?

3 volts?

 

[cnh1995]

3 volts?

This may be the final answer-voltage across each capacitor in steady state(edit:no this is not, I didn't read the question first!). What I mean by "net voltage" is the voltage responsible for circulating current. Since current is 0, net voltage would also be 0, wouldn't it? In fact, because the net voltage is 0, the current stops, doesn't it?

 

[jbriggs444]

3 volts?

The voltage across the charged capacitor started at 3 volts before it was [partially] discharged into the uncharged capacitor. 3 volts cannot be correct for the resulting state after all the transients have settled out.

 

[**Mariam**]

The voltage across the charged capacitor started at 3 volts before it was [partially] discharged into the uncharged capacitor. 3 volts cannot be correct for the resulting state after all the transients have settled out.

Ok, can I calculate the equivalence capacitance so it is : 21/10
And then by V=Q/C i find V=90/21?

Sorry but I am self-studying electricity for a coming exam and I don't understand circuits much.

 

[cnh1995]

Ok, can I calculate the equivalence capacitance so it is : 21/10
And then by V=Q/C i find V=90/21?

Sorry but I am self-studying electricity for a coming exam and I don't understand circuits much.

90/21? That turns out to be more than 4V. You had 3V initially. How would the voltage increase?

 

[**Mariam**]

90/21? That turns out to be more than 4V. You had 3V initially. How would the voltage increase?

Well now I'm just clueless :/

 

[jbriggs444]

Ok, can I calculate the equivalence capacitance so it is : 21/10
And then by V=Q/C i find V=90/21?

You need to show your work, justify your steps and keep track of the units. The first problem is your computation of the equivalent capacitance. There are two problems with the stated result of 21/10.

1. No units are shown.
2. The stated result was computed based on an assumption that the capacitors are in series.

Edit: [removed an erroneous third complaint]

The capacitors are connected end to end. It is easy to imagine this as a series circuit from one point back around to the same point. But that picture does no good.

Instead, imagine one node on the positive side of the charged capacitor and the about-to-be-positive side of the uncharged capacitor. Now imagine a second node on the negative and about-to-be-negative side of the capacitors. The capacitors connect these two nodes in parallel.

 

[cnh1995]

Well now I'm just clueless :/

First, you need to understand what is happening. You have a capacitor of 3nF charged to 3V. It will have some charge Q and it will retain the charge unless any discharge path is provided.Now, you provided a discharge path to it through 7nF capacitor. What do you think will happen now? Draw the circuit and see.

 

[cnh1995]

Ok, can I calculate the equivalence capacitance

Yes, that will work, but with a little modification. Jbriggs444 has pointed out a flaw in your assumption.

 

[**Mariam**]

[img=http://s11.postimg.org/vun2usr4v/image.jpg]

 

[cnh1995]

So how about this:

http://[url=http://postimg.org/image/vun2usr4v/][img=http://s11.postimg.org/vun2usr4v/image.jpg][/url]

I can't see the image.

 

[cnh1995]

Do you see what is happening during the discharge? What parameters will change and what will remain same?
Edit: I now can see the image. I think there's no discharge since battery is there.

 

[**Mariam**]

Do you see what is happening during the discharge? What parameters will change and what will remain same?

But here again the total voltage is higher than voltage in the 3C capacitor.
In fact if you add the answers I got for voltage they equal the (90/21) I mentioned before ??

 

[cnh1995]

Ok. So, there is a battery in the circuit and it is kept there throughout. I thought there are just two capacitors.

 

[**Mariam**]

Ok. So, there is a battery in the circuit and it is kept there throughout. I thought there are just two capacitors.

Then What does that indicate?

 

[cnh1995]

Then What does that indicate?

This indicates that the voltage across capacitors will be constant i.e.3V, no matter how many capacitors you add in parallel to the original 3nF capacitor. There's no "discharge" now.

 

[Ray Vickson]

[img=http://s11.postimg.org/vun2usr4v/image.jpg]

You have a battery in your circuit. The original problem mentioned noting about a battery, and in fact probably did not want one! If you go back and read your textbook, or look on-line for similar situations, you will see what you need to do. It really is NOT difficult, but you do need to study and know the basics. Just "guessing" will not be good enough.

 

[**Mariam**]

You have a battery in your circuit. The original problem mentioned noting about a battery, and in fact probably did not want one! If you go back and read your textbook, or look on-line for similar situations, you will see what you need to do. It really is NOT difficult, but you do need to study and know the basics. Just "guessing" will not be good enough.

Ok, I'll review these concepts and then I will try to resolve it.

 

[Ray Vickson]

Ok, I'll review these concepts and then I will try to resolve it.

Excellent. That is what I hoped you would say.

 

[CWatters]

Going back to your reply to my earlier post...

And I don't think the charge leaves the circuit.

Correct.

So for capacitor 1: Q=3*3= 9C
And capacitor 2: Q=0
Total charge= 9C

Correct

And if they reach equillibrium the two charge distributes, but is it equal?why?

No. There is nothing that forces the charge to end up equally distributed between the two capacitors.

The two capacitors are connected together in a way that means something else will be "equal" on both (even if it's not immediately obvious what the actual value is of the something).