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Capacitors: Explain result from experiment

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[SOLVED] Capacitors: Explain result from experiment

Homework Statement


I made an experiment today at school with a parallel plate capacitor. I induced 100 V and increased the distance between the plates. I measured the charge and calculated the capacitance of the capacitor, which of course got smaller when the distance got bigger.

Now the thing is: When I calculate the capacitance theoretically (using C = K*A/d), where A is the area of the plate, I get values for the capacitance, which are a lot smaller than the measured values when the distance is very "large". Why is that?

I would think it would have to be something like the plates are getting discharged the further they are apart.
 

Answers and Replies

  • #2
Hootenanny
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Erm, how did you measure the charge?

Edit: Also, how did you increase the distance of the plates?
 
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  • #3
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I measured the charge using a Coulomb-metre. I increased the plates using plastic-"feet". My teacher gave them to me.
 
  • #4
Hootenanny
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One possible explanation for the discrepancy between the theoretical and experimental values is edge effects. Your expression for the capacitance ignores edge effects and assume a completely uniform electric displacement field, which is fine when the separation is small. However, once d becomes large the edge effects become more significant and would reduce the magnitude of the electric field between the plates and hence increase the capacitance.
 
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What are edge-effects in this case?
 
  • #6
Hootenanny
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What are edge-effects in this case?
Your expression for capacitance assumes that the electric displacement field is completely uniform. However, in reality at the edges of the capacitor the field is non-uniform, the field lines arc as in the diagram below.

http://www.phys.unsw.edu.au/PHYS1169/beilby/capacitors_files/beilby_01_0001.gif [Broken]
 
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Ahh, I see. Yes that actually does make a lot of sense.. Of course my first assumptions (the plates getting discharged) is wrong, since nothing would discharge them.

I understand it. Thanks!

Edit: But how would that result in the experimental value being lower than the theoretical? Would in be the other way around?
 
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  • #8
Hootenanny
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Edit: But how would that result in the experimental value being lower than the theoretical? Would in be the other way around?
However, once d becomes large the edge effects become more significant and would reduce the magnitude of the electric field between the plates and hence increase the capacitance.
Hence the theoretical result (ignoring edge effects) would predict a lower capacitance than is observed experimentally.
 
  • #9
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Sorry, I really have a bad habit of just skating through a post, and asking all sorts of questions which are already answered in the post. I will stop that.

Thanks again!
 
  • #10
Hootenanny
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Sorry, I really have a bad habit of just skating through a post, and asking all sorts of questions which are already answered in the post. I will stop that.
Don't worry about it :smile:
 

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