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Electric field for a capacitor

  • Thread starter vysero
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  • #1
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Homework Statement



I have browsed similar threads and have only found more confusion. My question is if I have two parallel plates with an area A = 1m^2 and they are separated by a distance d = .03m. Each has a charge of Q and -Q respectively where Q = 1nC. I want to find the electric field between the plates.[/B]

Homework Equations



The equation I used was from my textbook: δ/2ε˳[/B]

The Attempt at a Solution



So I said that E(+) = 1nC/2(8.85x10^-12) and that E(tot) = 2(E(+)) = 112.99 n/c. However, this calculation does not take into account the distance between the two plates which I feel it should. Where have I gone wrong here?[/B]
 

Answers and Replies

  • #2
Svein
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You would need to calculate the capacitance - and then use Q = CV.
 
  • #3
gneill
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You can calculate the field by going the route suggested by Svein, or by using the formula provided. The distance between the plates doesn't matter for the second method provided that the geometry of the capacitor allows the "field is uniform between plates" assumption to hold and what's called "fringe effects" can be neglected. This is almost always the case for problems that are assigned.
 
  • #4
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You can calculate the field by going the route suggested by Svein, or by using the formula provided. The distance between the plates doesn't matter for the second method provided that the geometry of the capacitor allows the "field is uniform between plates" assumption to hold and what's called "fringe effects" can be neglected. This is almost always the case for problems that are assigned.
Okay so where did I go wrong when using the provided formula? Or did I do that right?
 
  • #5
gneill
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Looks right to me. You summed the field contributions of the two plates.
 
  • #6
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Looks right to me. You summed the field contributions of the two plates.
Oh :P okay thanks!
 

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