How Do You Solve Capacitor Equilibrium Problems?

[Regtic]

Homework Statement

Homework Equations

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

U = Q² / 2C
U = potential energy

The Attempt at a Solution

I'm just stuck on b. I tried doing Q_eq= √(2UC) then using the fact that q₁+q₂ = Q which is Q_eq 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 = Q_eq

But that didn't work.

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

 

[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?

 

[Regtic]

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?

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

 

[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?

 

[Regtic]

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?

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?

 

[SammyS]

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?

Of course charge is conserved.

Unless Q₁ = Q₂, Q₁ - Q₂ is not zero.

 

[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).

 

[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.

 

[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\muC which is off from what it should be.

Answer should be Q₂ is 250

 

[ehild]

So I tried

\frac {C_1 +C_2} {V_o} = Q_e

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

ehild

 

[SammyS]

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

So ...

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

 

[Regtic]

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

ehild

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\muC which is still wrong.

 

[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.

 

[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

 

[Regtic]

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

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?

 

[SammyS]

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?

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?

 

[Regtic]

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?

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

 

[SammyS]

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

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.

 

[Regtic]

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.

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

 

[SammyS]

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

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?

 

[Regtic]

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?

Hmmm so if C₂ had a higher charge or if the current went in the other direction because the plates swapped orientations, the charge would be negative?

 

[Simon Bridge]

Hmmm so if C2 had a higher charge or if the current went in the other direction because the plates swapped orientations, the charge would be negative?

Please reread post #7 in light of what you have learned.

Before they are connected, the charge on Capacitor 1 will be higher than the charge on Capacitor 2 because C1>C2 but the voltages are the same.

If C2>C1 then the opposite situation would prevail.
"swapped orientations" is the same as looking at the other side of the capacitors.

Note:
The net charge on each side stays the same.
It does not matter which plate is positive and which negative to answer the question.

 

[ehild]

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

Two plates are connected. Charge can migrate to one to the other, but nowhere else. The sum of the charges on C1 and C2 must be constant. Q1=1500 μC, Q2=-750 μC, so the whole charge on the connected upper plates is Q=750 μC. At the same time, there is -1500 + 750 = -750 μC charge on the bottom plates.

That charge is shared between the capacitors, the charge of C1 becomes q1 and the charge of C2 becomes q2, positive on the upper plates and negative, -q1 and -q2 on the bottom plates. You have found the common voltage already: U=Q/C_eq=(750/9 )= 83.33 V.
Using q=UC, find the individual charges.


ehild

 

[SammyS]

I was going at this from a somewhat different point of view than are ehild or Simon, so just follow them.

 

[Regtic]

Two plates are connected. Charge can migrate to one to the other, but nowhere else. The sum of the charges on C1 and C2 must be constant. Q1=1500 μC, Q2=-750 μC, so the whole charge on the connected upper plates is Q=750 μC. At the same time, there is -1500 + 750 = -750 μC charge on the bottom plates.

That charge is shared between the capacitors, the charge of C1 becomes q1 and the charge of C2 becomes q2, positive on the upper plates and negative, -q1 and -q2 on the bottom plates. You have found the common voltage already: U=Q/C_eq=(750/9 )= 83.33 V.
Using q=UC, find the individual charges.


ehild

Oh that makes a lot of sense. I sometimes become so dependent on the equations and specific phenomena that I have studied during the course that some of the most intuitive and simple problems suddenly become really complicated to me because they aren't described by a formula or aren't something I've seen before. I had my exam today and I did really well. Thanks for all your help, I appreciate it. I hope I can help others on here as you have helped me :)