Force between two point charges

  • Thread starter etown
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  • #1
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Homework Statement


Two positive point charges are 4.60 cm apart. If the electric potential energy is 75.0 x 10^-6 J, what is the magnitude of the force between the two charges?

Homework Equations


U=kq/r (1)
F=kqq/r^2 (2)


The Attempt at a Solution


So I tried using (1) to solve for q, and then i used the value i got for q in (2). the answer is supposed to be 1.63×10−3 N, but i'm just not getting it.
 

Answers and Replies

  • #2
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You are using [tex] \Phi = \frac{kq}{r} [/tex] where [tex] \Phi [/tex] is the electric potential.

It's important to note that the electric potential is not the same thing as the electric potential ENERGY. For two point charges, the electric potential energy is given by [tex] U = \frac{kq_1q_2}{r} [/tex] where r is the distance between the two charges.

That's the only reason you're getting the wrong answer.
 
  • #3
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thanks!
 
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