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Capacitance and electric fields

  • Thread starter Niles
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  • #1
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[SOLVED] Capacitance and electric fields

Homework Statement


I want to find the capacitance of a parallel-plate capacitor consisting of two metal surfaces og area A held a distance d apart.

First I find the electric field, which I know is [tex]\frac{\sigma }{{2\varepsilon }}[/tex]. Then I use that the potential is the line integral from minus-charge plate to positive-charge plate of electric field, so

[tex]V = \frac{\sigma }{{2\varepsilon }}d[/tex]

But in my book they do not divide by 2. This is because they have used superposition, which is OK - I agree with that.

Next, we consider the following:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capcyl.html#c2

Here they have used the electric field for |one| cylinder, they have not used superposition. This I do not agree with - we are still between the cylinders, so why do we not multiply by 2 here?
 
Last edited:

Answers and Replies

  • #2
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Oh, right.. because the charge inside our gaussian surface is the charge coming from the "inside-cylinder"...

I get it now. Thanks
 
  • #3
Actually the electric field is not between two cylinders, but it is in between a cylinder and a pipe, so it is not the same case as two parallel capacitors.
 

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