Oh, for some reason I thought the energy has to be positive, haha.
So then I get q_1 = -\frac{k r}{q_2}.
And V = -\frac{k^2}{q_2}?
(where k = \frac{1}{4\pi\epsilon_0})
Thanks for your help!
Homework Statement
A charged particle is fixed in place at the origin. A second particle of charge +10^{-6} C is released from rest from very far away (\approx (\infty, 0)). The second particle passes the point (9 m, 0) with a kinetic energy of 1.0 J.
Find the electric potential due to...