# Coulomb Potential Energy - discrepancy between like and opposite charges

1. Jun 3, 2010

### aalnaif

The Coulomb potential energy between two point charges is defined as:

V=[(q_1)(q_2)]/[(k*r)]

Suppose that you have two equal, like charges at a distance L, then V_like=q2/(k*L)

Similarly, for two equal, opposite charges, V_opp=-q2/(k*L)=-V_like

Both situations experience a force of equal magnitude (just opposite directions), yet V_opp<V_like? Shouldn't the two potential energies be equal?

By analogy with a mechanical spring, a weight that is left of the equilibrium position experiences a force of equal magnitude but opposite direction to a weight on the right of the equilibrium position. This is similar to the potential energy above. However, in this case, V_left=V_right, since the spring potential energy is:

V = 0.5kx2

2. Jun 3, 2010

### diazona

Force is actually related to the derivative of the potential energy. The actual value of the potential energy doesn't matter at all. If you take the derivative of V with respect to position, you'll see that for like charges, it is the opposite of the derivative for unlike charges. Same with the mechanical spring: the derivative of V is the opposite for the mass on the left as for the mass on the right.