- #1
KayDe
- 2
- 0
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/rf-1/ri)
[rf= final distance ; ri= 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