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Force of electric field and distance

  • Thread starter azurken
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  • #1
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Homework Statement


Two charges are placed between the plates of a parallel plate capacitor.
One charge is q1 and the other is q2 = 5.00 microC. The charge per unit
area on each of the plates has a magnitude of σ = 1.30 x 10^-4 C/m^2. The
magnitude of the force on q1 due to q2 equals the magnitude of the force
on q1 due to the electric field of the parallel plate capacitor. What is the
distance r between the two charges?

q2=5.00 x 10^-6
σ = 1.30 x 10^-4
r?

Homework Equations


E=σ/E0
E=F
F=8.99x10^9 x (q1 x q2)/r^2
E0=8.852x10^-12

The Attempt at a Solution


The only thing I don't know what to do with this one is the q1. I'm sure i'm using the right formulas but I can't figure out what to do with the q1.

I'm assuming my final answer will be something like r = sqrt((k x q1 x q2)/E)

The E which I can get from dividing the sigma by the E0 which should be 14685946.68



The final answer should be 5.53 cm but I honestly do not know how to get there because I don't know how to solve for q1.
 

Answers and Replies

  • #2
TSny
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Sometimes an unknown quantity will cancel out when you set up your equation(s). See if that's true for q1.

[Also, what's the meaning of your equation E = F ?]
 
  • #3
ehild
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Write out the the force exerted on q1 by q2 and also the force exerted on q1 by the electric field between the capacitor plates. The forces are equal.

ehild
 
  • #4
TSny
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I'm assuming my final answer will be something like r = sqrt((k x q1 x q2)/E)
That's almost correct. It looks like the thing that's throwing you off is your incorrect equation E = F. What is the correct equation for the force on a charge due to an electric field?
 
  • #5
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Alright yeah, so I think I can do

F=k(e^2)/r^2 which is the force of the capacitator on q1

which would then = to the same force of q2 exerted on q1

so Fq2 = Fcap but how do I find Fq2 now? I don't know the formula for electrons.

Can I just assume that q2 = -q1? and skip the Fq2 altogether?


edit: I mean if q2 and q1 have the same magnitudes but a different negativity.
 
  • #6
TSny
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F=k(e^2)/r^2 which is the force of the capacitator on q1
No, Coulomb's law is only for two point charges. To get the force of the capacitor on q1, use the idea that the capacitor creates an electric field, E, which you have already calculated. Then, you should be able to use E to calculate the force that E produces on q1. (There is a very important and basic relationship between E and F. I'm sure you've covered it.)
 

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