Electric Potential Energy Problem

[KayDe]

Homework Statement

Two Charged Objects (Q₁= -9.0nC & Q₂= 15nC) are fixed in place on the +x-axis. The first charge is 12 cm to the left from the origin the other is 12 cm to the right from the origin. A proton is moved from the origin to a position on the +y-axis 9.0 cm from the origin. a) What is the change in potential energy of the proton due to Q₁ & Q₂? [/B]

Homework Equations

As given in our Course

F=k |(q₁q₂)| / r²
E=k |q| / r
Potential Energy (uniform electric field) = qEΔd
Potential Energy (not in uniform electric field) = KQq(1/r_f-1/r_i)
[r_f= final distance ; r_i= initial distance]

The Attempt at a Solution

[/B]
I tried calculating the electric field at the origin between the two points due to the two charged objects. I figured I needed to calculate the electric field since there is not a uniform electric field with two different point charges present. With the electric field I then tried to calculate the electric potential energy since I know the distance and the charge of the proton. I then did the same for the final position and then calculated the change in the potential energy. The given solution for this problem is 1.44 x 10-17 J which I did not get. I either am approaching this the wrong way, understanding it wrong or calculating the wrong things but I get a different answer every time and I'm not sure what I'm missing . Thank You

 

[gneill]

You won't need the values of the electric field here. You just need the electric potential energy of the proton in its initial and final positions.

 

[KayDe]

You won't need the values of the electric field here. You just need the electric potential energy of the proton in its initial and final positions.

I tried that too but I'm still not getting the right solution :/

 

[gneill]

Can you show your work in detail? Start with the potential energy at the origin.