1. The problem statement, all variables and given/known data Two particles each with a charge of +3.00 μC are located on the x axis, with one particle at x = -0.80 m, and the other particle at x = +0.80 m. a) Determine the electric potential on the y axis at the point y = 0.60 m. b) What is the change in electric potential energy of the system if a third particle of charge q = - 3.00 μC is brought from infinity to the point on the y axis where y = 0.60 m? 2. Relevant equations Vp=KQ/r Vb-Va Pythagoras Theorem 3. The attempt at a solution First I draw the diagram. I place one charge at x=-0.80m and the second at x=0.80m. The point at which I want to calculate the potential energy is at (0,0.60m). I apply the Theorem of Pythagoras to get the radial distance from where the point charge is located to the point I want to find potential. r=Sqrt(x^2 +y^2) r=1m Since I have two charges, the Total potential is equal to the sum of both of the charges. Vtotal= V1 +V2. Since Both charges are equal and are the same distance apart to the point p, Vtotal=53940 V. This is my answer for part A. for part B, I want to find the potential difference Vb-Va. A charge comes into point P from infinity, So we say that at infinity it has zero potential. However, at P, the potential is -44950 V. Now, my problem is. Can I use my result from part a to answer part b. Meaning Va=53940 V. Since the charges are not moving from P (the ones that moved from the x-axis). Would my answer be Vb-Va=-44950-53940 Vb-Va=-98,890? Sorry for this beginner problem. I am not sure if I am understanding the concept of Potential due to Point Charges.