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Coulomb Potential Energy - discrepancy between like and opposite charges

  1. Jun 3, 2010 #1
    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. jcsd
  3. Jun 3, 2010 #2

    diazona

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    Homework Helper

    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.
     
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