# Homework Help: Speed of a charge

1. Feb 13, 2006

### wr1015

A point charge Q = +85.5 µC is held fixed at the origin. A second point charge, with mass m = 0.0526 kg and charge q = -2.73 µC, is placed at the location (0.323 m, 0).

(a) Find the electric potential energy of this system of charges.

(b) If the second charge is released from rest, what is its speed when it
reaches the point (0.125 m, 0)?

ok so i found part (a) -6.496597059 J but how do i calculate (b)?? i've tried subtracting the electric potential energy of the distance traveled from .323 to .125 (.198 m) from the electric potential energy of the system to find the change in electric potential energy. Then i divided that by my second charge (-2.73E-6) to get my change in potential so i could plug that into $$v= \sqrt{2q\Delta V/m}$$ and am getting wrong answers... any guess on where i'm going wrong??

Last edited: Feb 13, 2006
2. Feb 14, 2006

### Andrew Mason

$v = \sqrt{2U/m}$ where U is the potential energy change.
I get the change in potential energy to be 10.3 J which gives a speed of 19.8 m/sec.

AM

Last edited: Feb 14, 2006
3. Feb 14, 2006

### Staff: Mentor

Initially, the charges are separated by 0.323m and have the potential energy that you calculated in part (a). Finally, the charges are separated by 0.125m and have potential energy = ???. You need to calculate the final potential energy, and then the difference between the initial and final potential energies.

4. Feb 14, 2006

### wr1015

thank you all for your help