How Is Kinetic Energy Calculated for a Moving Charge?

[ttiger2k7]

[SOLVED] Kinetic Energy of a Charge

Homework Statement

A charged particle q, which is fixed in position, produces an electric potential of 3.10E2 V at a distance of 1 m away (the potential from q vanishes at infinity). If a particle of charge Q=.00500 C is released from rest at a distance of 7.00 m from q, what will be its kinetic energy after moving very far from q?

Homework Equations

V = Ed
E = Fq
KE = q(V_initial-V_final)

The Attempt at a Solution

Since V_initial is given at 310 V, I need to solve for V_final.

Since V = Ed, I solved for the electric field of Q:

E=kq/d^2

E=(9x10^9)(.005)/7^2
E=10285714.29

Then, since V=Ed
V = 10285714.29.

This is where I get confused because I don't think this is a realistic number since V initial was 3.10E2. Please help. I think all I need to do is find V_final to solve the problem, but I am not doing it right. Am I going about this the correct way?

SOLVED: Nevermind, figured it out.

 

[Nate810]

V_final had to be 0V since it is very far away from q, so the kinetic energy is equal to q(V_initial). KE = (Q)(V_initial) = (.005)(310) = 1.55J

 

[Moetasim]

I was using the wrong equation for V_final. It should be V_final = kQ/d. Plugging in the values, I get V_final = 0.01428571429 V. Then, using the equation KE = q(V_initial-V_final), I get KE = 1.42 J.