Understanding Capacitor Voltage and Capacity in a Circuit

[Eitan Levy]

Homework Statement

First, we connect a capacitor to a voltage source with an electric force of 60V.
After that we put another capacitor with a capacity of 0.01nF instead of the voltage source. As a result, the voltage on the first capacitor is down to 40V.
1. What is the voltage on the second capacitor?
2. What is the capacity of the first capacitor?

Homework Equations

The Attempt at a Solution

This is a very new topic for me. Why the voltage on the second capacitor is 40V and not 20V and how can we calculate the capacity? (Maybe it's 40V because of Kirchhoff's voltage law?)
Thanks in advance.

 

[.Scott]

Some of what is not explicitly stated:
The second capacitor is placed across the first - in parallel.
The second capacitor starts out discharged.

So the voltage is going to be proportional to the charge and inversely proportional to the total capacitance.
The charge does not change, so the capacitance must increase by 60/40, a factor of 1.5.

 

[Eitan Levy]

Some of what is not explicitly stated:
The second capacitor is placed across the first - in parallel.
The second capacitor starts out discharged.

So the voltage is going to be proportional to the charge and inversely proportional to the total capacitance.
The charge does not change, so the capacitance must increase by 60/40, a factor of 1.5.

Sorry, the second capacitor is not charged at first.
This is how it looks:

 

[BvU]

Hi,

Not nicely following the PF guidelines , this post is.

But I'll help you get started:

With resistors, you have V = I R
With resistors in parallel, the voltages over each of the resistors are identical.
With resistors in series, the current through each of the resistors is the same.

With capacitors, you have Q = C V
So if you charge your first capacitor, it carries Q₁ = C₁ V₁ where V₁ is known, C₁ is not and Q₁ isn't either.

If you then replace the voltage source by an (uncharged) capacitor, the charge will redistribute over the two capacitors until the voltages are the same (because the charge can't go anywhere else -- and the two capacitors are parallel).
You have one V₂ (40 V) and two C, one C₂known, one C₁ unknown. But you can write:
Q₂=C₁V₂ + C₂ V₂

Since the charge hasn't gone anywhere else, Q₂ = Q₁

Three equations, three unknowns. Bingo.

 

[Eitan Levy]

Some of what is not explicitly stated:
The second capacitor is placed across the first - in parallel.
The second capacitor starts out discharged.

So the voltage is going to be proportional to the charge and inversely proportional to the total capacitance.
The charge does not change, so the capacitance must increase by 60/40, a factor of 1.5.

This doesn't give the correct answer. The capacity is 0.02nF.

 

[BvU]

It does give the correct answer.

 

[BvU]

You have to reason: C₁ + C₂ = 1.5 C₁ with C₂ known

 

[Eitan Levy]

Hi,

Not nicely following the PF guidelines , this post is.

But I'll help you get started:

With resistors, you have V = I R
With resistors in parallel, the voltages over each of the resistors are identical.
With resistors in series, the current through each of the resistors is the same.

With capacitors, you have Q = C V
So if you charge your first capacitor, it carries Q₁ = C₁ V₁ where V₁ is known, C₁ is not and Q₁ isn't either.

If you then replace the voltage source by an (uncharged) capacitor, the charge will redistribute over the two capacitors until the voltages are the same (because the charge can't go anywhere else -- and the two capacitors are parallel).
You have one V₂ (40 V) and two C, one C₂known, one C₁ unknown. But you can write:
Q₂=C₁V₂ + C₂ V₂

Since the charge hasn't gone anywhere else, Q₂ = Q₁

Three equations, three unknowns. Bingo.

Thank you, but I still don't understand how you can say that Q₁=Q₂

 

[Eitan Levy]

Hi,

Not nicely following the PF guidelines , this post is.

But I'll help you get started:

With resistors, you have V = I R
With resistors in parallel, the voltages over each of the resistors are identical.
With resistors in series, the current through each of the resistors is the same.

With capacitors, you have Q = C V
So if you charge your first capacitor, it carries Q₁ = C₁ V₁ where V₁ is known, C₁ is not and Q₁ isn't either.

If you then replace the voltage source by an (uncharged) capacitor, the charge will redistribute over the two capacitors until the voltages are the same (because the charge can't go anywhere else -- and the two capacitors are parallel).
You have one V₂ (40 V) and two C, one C₂known, one C₁ unknown. But you can write:
Q₂=C₁V₂ + C₂ V₂

Since the charge hasn't gone anywhere else, Q₂ = Q₁

Three equations, three unknowns. Bingo.

Wait, do you mean that the current can only be like that? I don't understand.

 

[.Scott]

Wait, do you mean that the current can only be like that? I don't understand.

The 60V deposited a charge on C1. Then the 60V was disconnected - leaving that charge (and voltage) on C1.
Then C2 was connected to C1, allowing that fixed charge to be shared by the two capacitors.

 

[Eitan Levy]

The 60V deposited a charge on C1. Then the 60V was disconnected - leaving that charge (and voltage) on C1.
Then C2 was connected to C1, allowing that fixed charge to be shared by the two capacitors.

So unless I connect another voltage source, the sum of the charges stays the same?
Can you explain why?

 

[.Scott]

So unless I connect another voltage source, the sum of the charges stays the same?
Can you explain why?

Yes. That charge is an actual difference in the number of electrons and/or electron holes between one side of the capacitor and the other. When you connect the two capacitors, some of those electrons and electron holes travel to that second capacitor. But the total number (the net difference from one side to the other) does not change.

 

[Eitan Levy]

Yes. That charge is an actual difference in the number of electrons and/or electron holes between one side of the capacitor and the other. When you connect the two capacitors, some of those electrons and electron holes travel to that second capacitor. But the total number (the net difference from one side to the other) does not change.

And their movement is like that?

Thank you so much!

 

[.Scott]

And their movement is like that?

Close enough.
Remember that the voltage applied to C1 was positive and electrons are negative. So when C2 is applied, electrons actually travel from its positive side to C1, making C2 more positive and C1 less positive. In the same way that there would be a brief current between the positive sides of C1 and C2, there would also be current between the "ground" side of those capacitors.