Effect of Dielectric Material on Capacitor Voltages in a Circuit

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Homework Help Overview

The discussion revolves around the effects of inserting and removing a dielectric material in a capacitor circuit consisting of three identical capacitors (C1, C2, C3) with no internal resistance. Participants explore how the voltages across the capacitors are affected when the dielectric is introduced and subsequently removed after disconnecting the voltage source.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants question how the insertion of the dielectric affects the voltages and charges on the capacitors, particularly after the battery is disconnected. There are inquiries about the changes in capacitance and whether current will flow between capacitors under different conditions.

Discussion Status

The discussion is active, with participants providing insights and asking clarifying questions. Some have suggested calculating voltages before and after the dielectric is inserted, while others are exploring the implications of disconnecting the battery and replacing it with a wire. There is recognition of differing interpretations regarding the circuit's behavior after the voltage source is removed.

Contextual Notes

Participants note the ambiguity in the original problem statement regarding the removal of the voltage source, leading to discussions about whether the circuit is left open or shorted. There is also mention of the assumption that the total charge remains constant once the battery is disconnected.

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Homework Statement


upload_2018-1-1_16-53-59.png

C1=C2=C3=C
No internal resistance.
We wait for a while and the put a dielectric material in C3. Then after waiting again we remove the voltage source and after that we take this material out. How will the voltages on the capacitor be affected?

Homework Equations


CV=Q

The Attempt at a Solution


I just don't understand how to know what will change and how to know what will happen in the circuit after that. I know that the voltage on C3 will be affected, but from there how to know what will happen to the other capacitors? What would give us a balanced circuit?
 

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If you can calculate all the voltages before and after the dielectric is inserted, you can compare and see what is different. Once you know the voltages, you can find the charge on each capacitor. Once the battery is disconnected, will the capacitors lose their charges? What about after the dielectric is removed with the battery still disconnected? Remember, the total charge stays the same because it has nowhere to go after the battery is disconnected.
 
kuruman said:
If you can calculate all the voltages before and after the dielectric is inserted, you can compare and see what is different. Once you know the voltages, you can find the charge on each capacitor. Once the battery is disconnected, will the capacitors lose their charges? What about after the dielectric is removed with the battery still disconnected? Remember, the total charge stays the same because it has nowhere to go after the battery is disconnected.
I understand the total stays the same, but how am I supposed to know if each capacitor's charge will be changed? Maybe because the voltages are not identical on each capacitor, it will change? I have no idea how to figure this out.
 
What happens to the capacitance of C3 when the dielectric is inserted with the battery still connected? Does it change or does it stay the same?
 
kuruman said:
What happens to the capacitance of C3 when the dielectric is inserted with the battery still connected? Does it change or does it stay the same?
It changes, the voltage changes and it's charge stays the same.
 
How do these changes occur? What increases and what decreases? Can you find V3 and Q3 before the dielectric is inserted? What about after the dilectric is inserted?
 
kuruman said:
How do these changes occur? What increases and what decreases? Can you find V3 and Q3 before the dielectric is inserted? What about after the dilectric is inserted?
I can find them before. After I am not sure how to.
 
  • #10
Eitan Levy said:
I am aware of that but how can I know if current will flow between capacitors?
You mean after the dielectric is removed with the battery disconnected? Yes, current will flow so that the new charge on C3 is the same as on the parallel combination.
 
  • #11
kuruman said:
You mean after the dielectric is removed with the battery disconnected? Yes, current will flow so that the new charge on C3 is the same as on the parallel combination.
I thought it stays the same anyway?
 
  • #12
Eitan Levy said:
...we remove the voltage source...
Can you clarify this? When the voltage source is removed, is the circuit left open where the source was or is the source replaced by a short circuit (a wire) to close the gap?
 
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  • #13
gneill said:
Can you clarify this? When the voltage source is removed, is the circuit left open where the source was or is the source replaced by a short circuit (a wire) to close the gap?
It is replaced by a wire I believe.
 
  • #14
The answer will be different if the battery is replaced with a wire as opposed to when it is not.
 
  • #15
kuruman said:
The answer will be different if the battery is replaced with a wire as opposed to when it is not.
I understand , but it doesn't say specifically what happened, just that the voltage source was removed.
According to the answers only the voltage on C3 will change, maybe you can understand from this?
 
  • #16
Can anyone help me out with this? I just have an exam tomorrow so I am pretty nervous.
 
  • #17
Eitan Levy said:
I understand , but it doesn't say specifically what happened, just that the voltage source was removed.
According to the answers only the voltage on C3 will change, maybe you can understand from this?

If only the voltage on ##C_3## changes then when the battery is removed leaves open circuit. Because if it was shorted with wire then essentially as you can verify the circuit becomes 3 capacitors in parallel so if voltage in one changes then voltage in all changes since they all have same voltage.
 
  • #18
Delta² said:
If only the voltage on ##C_3## changes then when the battery is removed leaves open circuit. Because if it was shorted with wire then essentially as you can verify the circuit becomes 3 capacitors in parallel so if voltage in one changes then voltage in all changes since they all have same voltage.
So if we put a wire where we took out the voltage source, current would have flown right? (Because V3=2ε/3 and V12=ε/3).
 
  • #19
Eitan Levy said:
So if we put a wire where we took out the voltage source, current would have flown right? (Because V3=2ε/3 and V12=ε/3).
Yes. It might be instructive to look at the charges on the capacitors: Make a circuit drawing showing the distribution of charges on the capacitor plates in the instant before the battery-replacing wire is inserted.
 
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  • #20
Eitan Levy said:
So if we put a wire where we took out the voltage source, current would have flown right? (Because V3=2ε/3 and V12=ε/3).
yes and if the connecting wire has zero resistance the current that will flow it will be infinite but will last only for infinitesimal time dt. The total charge that will flow from/into the C_3 capacitor will be finite Idt=infinite * 0=finite for this case.
 
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  • #21
Appreciate the help, thanks a lot!
 
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