electrostatics triangle corner charges question

Question

Imagine an equilateral triangle. The top vertex has a positive charge. The bottom left vertex has a negative charge. The bottom right vertex has a positive charge. What happens to electric potential energy as the bottom right charge moves away? (All the charges have equal magnitudes)

Answer 1

The electric potential stored in a formation of charges can be represented by the total work required to bring the charges to their current position from infinity. Since the electric potential is zero at infinity, we can just bring in each charge individually and then see what its electric potential would be at its new position. Since there are no charges present when we bring in the first charge, it would require no work to bring it into position. For the second charge, you have to consider the potential caused by the first charge, which would be kq₁q₂/r₁₂. For the third charge you factor in both of the other charges and get kq₂q₃/r₂₃ + kq₁q₃/r₁₃. This means that the total potential stored in a given configuration of 3 charges is kq₁q₂/r₁₂+kq₂q₃/r₂₃ + kq₁q₃/r₁₃. If you make the bottom right charge q3, and set the distances to be equal, as they are in this formation, then the two terms containing q3 will cancel out as q1 = -q2. This means that as long as the distance between q3 and each charge remains equal, the potential energy will remain constant. However, if the charge is moved away in a direction that leads to those distances not being equal, the change in potential energy will shift in the sign of the term who's particle it ends up closer to, then shift back towards zero as the distance approaches infinity. If you need any more clarification, let me know.