Capacitors: Explain result from experiment

In summary, the conversation discusses an experiment with a parallel plate capacitor where the distance between the plates is increased. The theoretical and experimental values for the capacitance are compared, with the theoretical value being smaller due to ignoring edge effects. The concept of edge effects and their impact on the electric field between the plates is explained. The end result is that the experimental value for capacitance is higher than the theoretical value due to these edge effects.
  • #1
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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.
 
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  • #2
Erm, how did you measure the charge?

Edit: Also, how did you increase the distance of the plates?
 
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  • #3
I measured the charge using a Coulomb-metre. I increased the plates using plastic-"feet". My teacher gave them to me.
 
  • #4
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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  • #5
What are edge-effects in this case?
 
  • #6
Niles said:
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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  • #7
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
Niles said:
Edit: But how would that result in the experimental value being lower than the theoretical? Would in be the other way around?
Hootenanny said:
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
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
Niles said:
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:
 

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy in the form of an electric charge. It is made up of two conductive plates separated by an insulating material called a dielectric.

2. How do capacitors work?

Capacitors store electrical energy by creating an electric field between the two conductive plates. When a voltage is applied to the capacitor, one plate accumulates a positive charge and the other accumulates a negative charge. This creates an electrical potential difference between the plates, which allows the capacitor to store energy.

3. What factors affect the capacitance of a capacitor?

The capacitance of a capacitor is affected by the surface area of the plates, the distance between the plates, and the type of dielectric material used. A larger surface area and smaller distance between the plates result in a higher capacitance, while using a dielectric with a higher permittivity can also increase the capacitance.

4. How do you measure the capacitance of a capacitor?

The capacitance of a capacitor can be measured using a capacitance meter or by using the formula C = Q/V, where C is the capacitance in farads, Q is the charge stored on the capacitor in coulombs, and V is the voltage across the capacitor in volts.

5. What can affect the accuracy of capacitance measurements in an experiment?

The accuracy of capacitance measurements can be affected by factors such as stray capacitance, which is unwanted capacitance that can influence the measurement, and the frequency of the applied voltage. It is important to use proper techniques and equipment to minimize these effects in order to obtain accurate results.

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