Print ViewElectric Potential Energy of Three Point Charges

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Homework Statement



Three equal point charges, each with charge 1.00 microC, are placed at the vertices of an equilateral triangle whose sides are of length 0.200 m. What is the electric potential energy U of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)
Use epsilon_0 = 8.85×10−12 F/m for the permittivity of free space.

Homework Equations



This is what I'm not sure of. I thought in order to find electric potential energy, you would multiply q by the distance and then add them together. (1xe-6)*(.200) =2e-7*(3) = 6e*-7 but that's wrong.



The Attempt at a Solution


see above.
 

Answers and Replies

  • #2
rl.bhat
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Electric potential energy = k*q1*q2/d.
 
  • #3
Spinnor
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Say the three charges are at infinity. Move one to one corner of your triangle. No work done, right? Now move the second one to another point of the triangle. For this configuration you should be able to write down the potential energy. To this potential add the work done moving the third charge in some symmetric way to the final point of the triange. Remember work equals the integral of F dot dx.
 
  • #4
Spinnor
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Say the three charges are at infinity. Move one to one corner of your triangle. No work done, right? Now move the second one to another point of the triangle. For this configuration you should be able to write down the potential energy. To this potential add the work done moving the third charge in some symmetric way to the final point of the triange. Remember work equals the integral of F dot dx.
The last step here while doable is the hard way to go about this.
 
  • #5
Spinnor
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The last step here while doable is the hard way to go about this.
Reading from Halliday and Resnick:

"The total potential energy U of the three-charge system is the sum of the potential energies associated with the three pairs of charges."

they do your problem as an example.
 
  • #6
tiny-tim
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Although moving the third charge means doing work against the force from the midpoint of the first two charges, which involves a cosine, so it looks as if you can't deal with each of the first two charges as if the other wasn't there (and then add), that doesn't matter, and you can, because …

W12 = ∫ (F1 + F2) . ds

= ∫ F1 . ds + ∫ F2 . ds

= W1 + W2 :smile:
 

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