Calculate Work Needed to Remove Capacitor Plate at 100V | Physics Homework

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Homework Help Overview

The problem involves calculating the work needed to remove a metal plate inserted between the electrodes of a capacitor, with considerations for both connected and disconnected scenarios at a voltage of 100V. The subject area includes concepts related to capacitance, energy stored in capacitors, and the effects of inserting a dielectric material.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the change in capacitance when the metal plate is inserted versus when it is removed. There are questions about the configuration of capacitors formed by the plate and how they are connected. Some participants suggest calculating capacitance and energy differences to determine the work done.

Discussion Status

Participants are exploring the relationships between capacitance, energy, and work done in the context of the problem. There is an acknowledgment of the need to calculate the resultant capacitance and energy differences, with some guidance provided on the formulas involved. Multiple interpretations of the setup are being discussed.

Contextual Notes

The problem includes specific parameters such as the thickness of the plate, surface area of the electrodes, and the distance between them, which may influence the calculations. Participants are also considering the implications of the capacitor being connected or disconnected from a voltage source.

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Homework Statement



A metal plate, which is d1=0.5mm in thickness, is put in parallel between the electrodes of a capacitor. Surface area of capacitor's electrodes is S=100cm2 and distance d between them is 2mm. What is the work that is needed to be done to remove the metal plate when capacitor is: a) disconnected from the source; b) connected to the source? The voltage of this source is U=100V.

Homework Equations



I believe it has something to do with the capacitor's energy formula W=CU2/2 and maybe Coloumb's Law.

I would be glad if someone could help me with this problem!
 
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What do you think about the change of capacitance when the plate is inside and outside?

ehild
 
ehild said:
What do you think about the change of capacitance when the plate is inside and outside?

ehild

I think that capacitance will be greater when the plate is inside rather than when it's outside. Am I right?
 
If you put a metal plate between two other ones then you have to capacitors instead of one. What is the capacitance of each of them? How are they connected?

ehild
 

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Noirro said:
I think that capacitance will be greater when the plate is inside rather than when it's outside. Am I right?

Do a calculation and find out! :smile:
 
Then they are connected in series. I think each one of those two capacitors has a capacitance of [tex]C=\frac{2\epsilon_{0}S}{d-d_{1}}[/tex]
 
Yes, correct! And what is the resultant capacitance of these two series connected ones?

ehild
 
Great! The resultant capacitance should be [tex]C=\frac{4\epsilon_{0}S}{d-d_{1}}[/tex]
 
Yes! Now the energy comes.

ehild
 
  • #10
I believe the energy is now [tex]W=\frac{2\epsilon_{0}S}{d-d_{1}}U^{2}[/tex]
Then maybe the work that is to be done is the difference of energy that the capacitor contains with and without the inserted plate?
 
  • #11
Yes, the difference of the energy is equal to the work done on the capacitor. If it is disconnected from the battery after it has been charged, the only work is done by the person who pulls out the plate. In this case, the charge stays the same on the plates. If the plates are connected to the battery, the voltage is unchanged and the charge will be different. Both the man and the battery does work.

ehild
 
  • #12
ehild said:
Yes, the difference of the energy is equal to the work done on the capacitor. If it is disconnected from the battery after it has been charged, the only work is done by the person who pulls out the plate. In this case, the charge stays the same on the plates. If the plates are connected to the battery, the voltage is unchanged and the charge will be different. Both the man and the battery does work.

ehild

I understand now. Thank you very much!
 
  • #13
Noirro said:
I understand now. Thank you very much!

Noirro -- please check your PMs...
 

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