# Capacitor homework question

1. May 13, 2014

### Regtic

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

2. Relevant equations

C =Q/V
c = capacitence
q = charge
v = potential difference

U = Q2 / 2C
U = potential energy

3. The attempt at a solution

I'm just stuck on b. I tried doing Qeq= √(2UC) then using the fact that q1+q2 = Q which is Qeq since charge should be conserved. Then from the fact that at equilibrium the potentials should be equal, q1$/$c1 =q2$/$c2

so

q2 + q2c1$/$c2 = Qeq

But that didn't work.

Apparently the answer is A) 281mJ B) Q1' = 500$\mu$C Q2' = 250$\mu$C C) V1' = V2' = 83.3V D) U' = 31.3mJ

Last edited: May 14, 2014
2. May 13, 2014

### Simon Bridge

Thing is - some of the charge is negative and some positive.
When the capacitors are joined - they become one equivalent capacitor with a capacitance you know how to find.

Concentrate on what happens to the charge on each plate, individually.
What happens when a positively charged plate is connected to a negatively charged one?

3. May 13, 2014

### Regtic

I would think that the charge on the positively charged plates would flow to the negatively charged plates to decrease their potential until the potential on both plates are even. So

$$\frac{Q_1}{C_1 } = \frac {Q_2}{C_2}$$

Don't have Q1 or Q2 though

Last edited: May 13, 2014
4. May 13, 2014

### Simon Bridge

The voltages start out equal - but opposite.
When the caps are separate, you know the voltage and the capacitance for each one - therefore you know how much charge is on each plate or each capacitor.

If you start out with one conductor with charge +Q1 and another conductor with charge -Q2, then bring the two conductors into contact, what is the final of the combined conductors?

5. May 14, 2014

### Regtic

c1v doesn't equal q1 though. they have a different voltage now that we don't know.

The final would be 0 because they are equal... are you trying to say charge isn't conserved because it gets neutralised?

6. May 14, 2014

### SammyS

Staff Emeritus
Of course charge is conserved.

Unless Q1 = Q2, Q1 - Q2 is not zero.

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7. May 14, 2014

### Simon Bridge

Thank's SammyS.
That is correct, there is no reason to assume that the charge on each plate is the same magnitude.
Before connection: Q1=C1V0, Q2=C2V0, the only way Q1=Q2 is if C1=C2.

After the connection you have an effective capacitance C with a total net charge Q on each plate.
Once you know how much charge is distributed between the capacitors, it remains to work out how it is distributed.

That gives you the answer to (b).

8. May 14, 2014

### rude man

Another approach is to connect the capacitors thru a resistor of any value, then solve for the two voltages for t = infinity. Of course, they will be the same.

9. May 14, 2014

### Regtic

So I tried

$$(C_1 +C_2)V_o = Q_e$$
since they are both charged by individual batteries before being connected
$$\frac {C_1 Q_2} {C_2} = Q_1$$
and
$$Q_1 +Q_2 = Q_e$$
so
$$\frac {Q_e} {1+ \frac {C_1} {C_2} } = Q_2$$

That gave me Q2 = 750$\mu$C which is off from what it should be.

Answer should be Q2 is 250

Last edited: May 14, 2014
10. May 14, 2014

### ehild

That is wrong. Q=CV and not C/V.

ehild

11. May 14, 2014

### SammyS

Staff Emeritus
So ...

The first thing is to determine Q1 and Q2 , the charge on each capacitor prior to the switches being closed.

12. May 14, 2014

### Regtic

Woops, meant to write the opposite. I actually did that method last night and just tried to redo it quick to understand why it was wrong. Fixed it but my answer becomes 750$\mu$C which is still wrong.

13. May 14, 2014

### Regtic

Am I wrong or is the answer package wrong you think? I don't think it's possible that Q2 isn't 750 micro coulombs. Answer package says q1 is 500 and q2 250 and that would be impossible if charge is conserved. Either that or my Qeq is wrong which I don't think it is.

14. May 14, 2014

### ehild

Initially you have 1500 μC on C1 and 750 μC on C2. The capacitors are connected, positive plate to the negative one. The charge can migrate between the connected plates, and 1500 μC - 750 μC = 750 μC - that is the charge of the resultant equivalent capacitor; and Ceq=9μF.
From these data you get the voltage, which is equal to the common voltage across the individual capacitors. You get the new charges on both capacitors from the common voltage and the capacitance.

ehild

Last edited: May 14, 2014
15. May 14, 2014

### Regtic

OOOOOOOH. Because they are connected from positive to negative, the resultant net charge is no longer 750+1500 it's 1500-750 because the charge migrates. But why is it 1500-750 and not 750-1500? Could have just as easily been negative no? Is it just because we had a positive voltage that we know the charge is positive?

16. May 14, 2014

### SammyS

Staff Emeritus
Of course it's true, that the charge can be negative.

Capacitors in general have some charge, q, on one plate and the opposite, namely -q, on the other.

With the figure you show in the OP, what sign is the charge on the upper plates after the switches are closed?

17. May 14, 2014

### Regtic

i's still going to be positive on the left and negative on the right assuming the resultant net charge is positive.

18. May 14, 2014

### SammyS

Staff Emeritus
Once they're connected, they share the same voltage, with the same polarity.

Opposite charges do attract and being connected with a conductor will allow that to happen to the max.

19. May 14, 2014

### Regtic

I don't get it, how do you know that the charge is positive?

20. May 14, 2014

### SammyS

Staff Emeritus
Initially, what's the charge on the upper plate of C1. (The figure shows it to be positive.)

Initially, what's the charge on the upper plate of C2. (The figure shows it to be negative.)

Which will be more than enough to neutralize the other?