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Negative potential energy

  1. Aug 12, 2008 #1
    The coulomb force is given by k*q1*q2/r^2 so if there are two charges that have negative signs, then the force would be negative. Having said that, a "negative force" would then signal an attractive force since opposites attract.

    F=-dU/dx, so a negative force would result in a "positive potential"

    Does this mean a positive potential is attractive and a negative potential is repulsive? I'm not sure why there is a negative sign there. If this is the case, then why is the binding energy for an electron around the hydrogen atom -13.6 eV? The energy here is negatice, but the force is "binding"

    I'm confused, any thoughts would be appreciated.
     
  2. jcsd
  3. Aug 12, 2008 #2
    The + and - simply mean how the system is bound. For example, the - sign in gravity simply implies that unless one piece of matter is infinitely far from the others, that particle remains bound to the others; however, the gravitational force between matter is naturally attractive and so the gravitational potential energy is naturally bound to values increasing in magnitude below 0J. A + sign implies that the system is not bound, meaning that if 2 charges repel, the electric potential energy draws closer to 0J naturally.
     
  4. Aug 12, 2008 #3
    so how does force play into this? an attractive (negative) force results in a unbounded (positive) potential?
     
  5. Aug 12, 2008 #4

    Doc Al

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    Not a positive potential necessarily, but a positive rate of change of potential. Meaning: If the force points to the left, then the potential energy increases to the right.
    If, as the particles are separated, the potential energy increases, then the force between them is attractive.
    Realize that the zero point for potential energy is when the electron is far away from the nucleus. Compared to that point, the ground state electron PE is -13.6 eV. Thus, to move the electron from the atom requires energy to be added. (And thus the force is attractive, since moving the electron further out increases the potential energy.)
     
  6. Aug 12, 2008 #5
    Say we have two equal charges of equal mass and opposite sign. Charge A is -q and charge B is q, with the potential =0 and charge A at rest and charge B has an initial velocity towards A of v. So, as charge A goes towards charge B, then potential energy becomes NEGATIVE since it takes energy to pull them apart... however, this is counteracted by the increaseing kinetic energy... thus the total energy of the system will always be 0.5mv^2 right? (ignoring effects of radiation)
     
  7. Aug 12, 2008 #6

    Doc Al

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    Sure. As the charges get closer their potential energy decreases while their kinetic energy increases. The total energy remains constant.
     
  8. Aug 12, 2008 #7
    and by decrease, you mean "increase negatively" just like how the electron has a potential of -13.6eV

    does this mean the electron around the hydrogen atom can never have a positive potential?
     
  9. Aug 12, 2008 #8

    Doc Al

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    Sure, if you want to express it like that. But realize that it's not the sign of the potential that matters but how it changes.
    Right. As long as you take the usual convention of electric PE being zero at infinity, the electron in a hydrogen atom can never have a positive potential.
     
  10. Aug 12, 2008 #9
    If the particles are separated, the potential energy increses, then dU/dx is always positive, thus the force is always negative, then does this mean a negative force is always attractive?


    In this case, if the electron was moved from -13.6 to infinity, then the change in potential energy is positive, but this makes the force negative... does this mean that a negative force increases a potential (instead of saying a negative force means an attractive force)
     
  11. Aug 12, 2008 #10

    Doc Al

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    It depends on what you mean by "negative"--that's a sign convention.

    Let x represent the separation of particle B with respect to particle A. As x increases, B gets further away. Since dU/dx is positive, the force is negative and thus points in the negative x direction. That means the force on B points towards A, and thus is an attractive force.
    Read what I just wrote above and rephrase the question if I missed it.
     
  12. Aug 12, 2008 #11
    the coulomb force is attractive between the two charges, but why is the force that is needed to move charge B away not taken into account?

    also, what if we let the charge B move towards charge A, then since the potential is decreasing, dU/dx<0, making the force positive?
     
  13. Aug 12, 2008 #12

    Doc Al

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    Not sure what you mean. The force needed to (just) move charge B away is equal and opposite to the Coulomb force. (Assuming charge A is fixed.)

    F = -dU/dx gives you the force associated with the potential. Any other force is up to you.
    dU/dx is still > 0. The slope of U, and the associated force, doesn't change just because the charge is moving one way or the other.
     
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