# 2 capacitors reaching equillibrium?

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1. Jan 8, 2016

### **Mariam**

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Q=CV

3. 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?

2. Jan 8, 2016

### CWatters

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?

3. Jan 8, 2016

### **Mariam**

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?

4. Jan 8, 2016

### jbriggs444

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?

5. Jan 8, 2016

### **Mariam**

Is potential difference 0?

6. Jan 8, 2016

### 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.

7. Jan 11, 2016

### **Mariam**

Oh ok.

But I still don't get the question

8. Jan 11, 2016

### cnh1995

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

9. Jan 11, 2016

### **Mariam**

3 volts?

10. Jan 11, 2016

### cnh1995

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?

11. Jan 11, 2016

### jbriggs444

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.

12. Jan 11, 2016

### **Mariam**

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.

13. Jan 11, 2016

### cnh1995

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

14. Jan 11, 2016

### **Mariam**

Well now I'm just clueless :/

15. Jan 11, 2016

### jbriggs444

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.

16. Jan 11, 2016

### cnh1995

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.

17. Jan 11, 2016

### cnh1995

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

18. Jan 11, 2016

### **Mariam**

19. Jan 11, 2016

### cnh1995

I can't see the image.

Last edited by a moderator: May 7, 2017
20. Jan 11, 2016

### 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.