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Electric Potential Energy of A System of Point Charges!

  • Thread starter Oaksmack
  • Start date
  • #1
A Particle of charge q=7.5x10^-6 Coulombs is released from rest at .6m on the x axis. The particle moves due to charge Q = -20x10-6 C. What is the Kinetic Energy of the particle the instant it has moved .4 meters if Q stays fixed at the origin?

Here I used KE1 + U1 = KE2 + U2
Which equals 0 + (kqQ)/.6 = KE2 + kqQ/.2
I got 2.2475 = KE2 - 6.7425
so KE2 = 8.99 or 9 Joules.
The answer is 4.5 J in the book however.

Where did I go wrong? My answer for the first part of the question, with Q =positive 20 micro Coulombs, was correct. Is the KE for attraction supposed to be multiplied by 1/2 as opposed to repulsion? Please help!

My k value by the way is correct, at 8.99x10^9.

Answers and Replies

  • #2
both potential energies should have the same sign. In your equation, one of them is positive the other is negative.