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

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The discussion focuses on calculating the work required to remove a metal plate from a capacitor under two conditions: when disconnected from the source and when connected to a 100V source. Participants explore how the presence of the plate affects capacitance, concluding that capacitance increases when the plate is inserted. The energy stored in the capacitor is derived from the formula W=CU²/2, and the work done is determined by the difference in energy with and without the plate. When disconnected, only the person removing the plate does work, while both the person and the battery do work when connected. The conversation concludes with a clear understanding of the concepts involved.
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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 C=\frac{2\epsilon_{0}S}{d-d_{1}}
 
Yes, correct! And what is the resultant capacitance of these two series connected ones?

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

ehild
 
  • #10
I believe the energy is now W=\frac{2\epsilon_{0}S}{d-d_{1}}U^{2}
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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