Circuits with capacitators connected parallel

[mbkau00]

C1 has capacitance of 0.05 μF
C2 has capacitance of 0.1 μF

C1 fully charged, connected to a 0.4 V battery
C2 left uncharged.

C1 was disconnected from the battery and two parts joined, allowing the charge to transfer from C1 to C2.

Q:
Calculate the percentage of charge transferred from C1 to C2.


I've calculated the charge on C1, from using Q = CV, as 2x10^-8 C

I'm unsure how to calculate C2 or a percentage of charge transfer.

Thanks in advance.

 

[BruceW]

Just think of it as a circuit with two capacitors (which each have a potential difference across them). What Law would you use to find the relationship of the potential difference across each capacitor? Or in other words, what Laws do you know about circuits that might be useful?

 

[mbkau00]

Hi, thanks for the reply! I'm very new to physics

I know that a capacitor in parallel has a total capacitance of C = C1 + C2.

Also know Kirchoff's laws, but i don't think they relate here? - as they pertain to currents entering and leaving a junction, or total EMF regarding Voltages.

 

[BruceW]

Yes, Kirchhoff's laws are the key! (They often are in circuit questions). The one about total EMF regarding Voltages is important for this question. Think about the capacitors after they have gotten into equilibrium (you don't need to do anything complicated like integrate the current).

 

[mbkau00]

Thanks for your help.

My understanding of Kirchoff's law relating to EMF is: The sum of all voltage changes must = 0 in a closed loop system? I still don't know how this helps.

Is there another formula I need to calculate the current? There are no resistors in this circuit.

I would have thought that once the circuit was reconnected, the charge on C1, 2x10^-8C would be distributed throughout so C2 would be charged to its full capacitance? Once finding the capacitance of C2, take that as a % of C1.

for C2: It isn't connected to a battery so I don't think I can use Q=CV as there is no voltage component? This is where I'm stuck in calculating anything to do with C2 to be able to relate to C1
All I have iss C2 = 0.1x10^(-6) F

Thanks so much again.

 

[gneill]

After C1 is disconnected from the battery it has a pool of charges stored on its plates (as determined by the formula Q = C*V). These charges are available to move if the capacitor gets connected to something else and forms a circuit (closed path).

In this case the available charges will end up distributed across both capacitors. The trick is to find out how much of the charge will end up on each capacitor.

Since the capacitors end up connected in parallel, what can you say about the final potentials of both capacitors? Will any charge go missing?

 

[mbkau00]

Thanks for your reply

Yes I understand I need to find out how much of the available charge will end up on each. The answer given is 70% and I have no idea how. I don't know how or where to use the known charge on C1.


C1 has a capacitance of 0.05uF, whereas C2 has a capacitance of 0.1uF. The transfer is from C1 to C2. Due to the capacitance differences, it can't be 100% transfer.

Capacitors in parallel have a combined capacitance of C = C1 + C2

 

[gneill]

Thanks for your reply

Yes I understand I need to find out how much of the available charge will end up on each. The answer given is 70% and I have no idea how. I don't know how or where to use the known charge on C1.


C1 has a capacitance of 0.05uF, whereas C2 has a capacitance of 0.1uF. The transfer is from C1 to C2. Due to the capacitance differences, it can't be 100% transfer.

Capacitors in parallel have a combined capacitance of C = C1 + C2

And the potentials?...

 

[mbkau00]

Sorry I'm unsure what or how I work a potential out.

The only formulae I know for capacitors are working out combined capacitance and the charge.

 

[gneill]

Sorry I'm unsure what or how I work a potential out.

The only formulae I know for capacitors are working out combined capacitance and the charge.

Two capacitors connected in parallel. Are their potentials the same or different?

 

[mbkau00]

I'm sorry I don't understand what a potential is.

If potential means capacitance, then no both capacitors have different capacitance.

 

[gneill]

I'm sorry I don't understand what a potential is.

If potential means capacitance, then no both capacitors have different capacitance.

Potential difference. Voltage as measured across the component.

 

[mbkau00]

I honestly don't know.

Having a guess i'll say there is a voltage difference between the two capacitors.

 

[gneill]

I honestly don't know.

Having a guess i'll say there is a voltage difference between the two capacitors.

I'll give you another guess... No, two components in parallel ALWAYS have the same potential difference across them. In this case they start out with different potentials (0.4V on C1 and 0V on C2), but as soon as they are connected they will quickly equilibrate to a common potential difference as the charges get redistributed. This redistribution will take very little time because there's (theoretically in this case) no resistance to impede the movement of charges.

Above you wrote the equation Q = CV for the charge stored on a capacitor. If V is now the same for both capacitors C1 and C2, then can you write equations for the charges Q1 and Q2 on the capacitors? What can you say about the sum of Q1 and Q2?

 

[mbkau00]

Thanks for that. So I'm assuming if they were in series, they would have diffreent potential differences?

How can I calculate what the new equilibrated charge is for the potential difference? Upon reconnection, I'd think 0.4V would be distributed across the two capacitators to a lesser voltage? If they are to have the same potential difference, is it .5*.4V ?

Q1 = 0.05uF * (.5*.4)
Q2 = 0.1uF * (.5*.4)

I don't know what the sum would equal.

 

[gneill]

Thanks for that. So I'm assuming if they were in series, they would have diffreent potential differences?

They probably would, yes.
quote]
How can I calculate what the new equilibrated charge is for the potential difference? Upon reconnection, I'd think 0.4V would be distributed across the two capacitators to a lesser voltage? If they are to have the same potential difference, is it .5*.4V ?
[/quote]
No, voltage is not distributed. Voltage is the result of charges on the capacitors. It's charge that gets distributed until the voltages on each capacitor are the same.

Q1 = 0.05uF * (.5*.4)
Q2 = 0.1uF * (.5*.4)

I don't know what the sum would equal.

Work symbolically until the end. You've got capacitances C1 and C2 with charges Q1 and Q2 and equal voltage V. Write an equation that describes equal voltages (using charges on each capacitor).

Regarding the sum of the charges, what charge are you starting out with? Do you expect charge to evaporate mysteriously or remain constant? Where would it go?

 

[mbkau00]

The only charge I know is on C1, which is 2x10^(-8)C.

2x10^(-8) = 0.05uF * .4V
Q2 = 0.1uF * .4V

 

[gneill]

The only charge I know is on C1, which is 2x10^(-8)C.

2x10^(-8) = 0.05uF * .4V
Q2 = 0.1uF * .4V

The final voltage is unknown at this point. It won't be the original voltage. You have only the total charge Q (= 2x10-8C) and the capacitances C1 and C2 to work with. Write an expression using (unknowns) Q1 and Q2 that equates the voltages on each capacitor. Then write another equation that relates the sum of Q1 and Q2. Two equations, two unknowns.

 

[mbkau00]

2x10^-8 = 0.05uF * V1
2x10^-8 = 0.1uF * V2

 

[gneill]

2x10^-8 = 0.05uF * V1
2x10^-8 = 0.1uF * V2

No, the final voltages must be the same for parallel connection. And the charges are currently unknown. Write the equations using variables, not constants.

 

[mbkau00]

C1 = 0.05uF*V
C2 = 0.1uF*V

 

[gneill]

C1 = 0.05uF*V
C2 = 0.1uF*V

I think you meant:

Q1 = C1*V
Q2 = C2*V

Solve them both for V and equate the results.

 

[mbkau00]

sorry yes,
V=Q1/C1
V=Q2/C2

equate? q1/c1 = q2/c2 ?

 

[gneill]

sorry yes,
V=Q1/C1
V=Q2/C2

equate? q1/c1 = q2/c2 ?

Yes. Now what do you know about the sum of q1 and q2?

 

[mbkau00]

sorry i don't know what the sum of charges equal.
q1+q2 = ?

 

[gneill]

sorry i don't know what the sum of charges equal.
q1+q2 = ?

What is the only source of charge available?

 

[mbkau00]

the charge on q1?
2x10^-8C

q1+q2 = 2x10^(-8)C

 

[gneill]

the charge on q1?
2x10^-8C

q1+q2 = 2x10^(-8)C

Yup

So now you have two equations in two unknowns. You can solve for q1 and q2.

 

[mbkau00]

The equations being?
V=Q1/C1
V=Q2/C2
q1+q2= 2x10^(-8)

 

[gneill]

The equations being?
V=Q1/C1
V=Q2/C2
q1+q2= 2x10^(-8)

Yes. Solve for q1 and q2 (and V if you need it).

 

[mbkau00]

doesn't V need to be solved for? substituting q1 into the last equation and solving for q2 will equal 0.

so to clarify:

c1 = 0.05uF
c2 = 0.1uF
q1 = 2x10^-8

V is unknown
q2 is unknown

 

[gneill]

doesn't V need to be solved for? substituting q1 into the last equation and solving for q2 will equal 0.

so to clarify:

c1 = 0.05uF √
c2 = 0.1uF √
q1 = 2x10^-8 X
Qtotal = 2x10^-8

V is unknown
q1 is unknown
q2 is unknown

You do not know what q1 will be when the capacitors are connected. You only know what it is BEFORE they are connected, and it happens to be the total charge available. q1 and q2 are your unknowns that represent the charges AFTER connection.

 

[mbkau00]

rearranging:

V=Q1/C1
V=Q2/C2
q1+q2= 2x10^(-8)

rearranging and substituting:

V=q1/0.05uF, 0.05uF*V=q1
V=q2/0.1uF, 0.1uF*V=q2

0.05uF*V+0.1uF*V = 2x10^-8

 

[gneill]

rearranging:

V=Q1/C1
V=Q2/C2
q1+q2= 2x10^(-8)

rearranging and substituting:

V=q1/0.05uF, 0.05uF*V=q1
V=q2/0.1uF, 0.1uF*V=q2

0.05uF*V+0.1uF*V = 2x10^-8

I'm not sure why you keep bringing V back into play. You're looking for the percentage of the charge that ends up being transferred from C1 to C2. So you only need the charges Q1 and Q2.

Eliminate V as you did earlier:

(1) Q1/C1 = Q2/C2
(2) Q1 + Q2 = Q

From (2): Q2 = Q - Q1

Replace Q2 in (1): Q1/C1 = (Q - Q1)/C2

solve for Q1.

Do the same for Q2. You should do this with the symbols because dragging around a lot of numbers through a derivation is both tiring and prone to error. Also, when done, you'll find that the equations are particularly elegant and easy to remember for similar problems in the future.

 

[mbkau00]

Thanks for your help and persistence.

Unfortunately I don't get an answer near 70% which is the correct answer. I think i must admit defeat, considering time spent.
Thanks again :)