Force between two point charges

AI Thread Summary
To find the force between two positive point charges separated by 4.60 cm with an electric potential energy of 75.0 x 10^-6 J, the relevant equations are U = kq/r for potential energy and F = kqq/r^2 for force. A common mistake is confusing electric potential with electric potential energy, which are distinct concepts. The correct approach involves using the potential energy equation to determine the charge value, which can then be substituted into the force equation. The expected answer for the force is 1.63×10−3 N, highlighting the importance of correctly applying the formulas.
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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.
 
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You are using \Phi = \frac{kq}{r} where \Phi 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 U = \frac{kq_1q_2}{r} where r is the distance between the two charges.

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