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Homework Help: Regarding the Electric Potential Formula

  1. Aug 9, 2014 #1

    The formula for Electric Potential as I have it is: "Let Q be a point source charge. At any point P that's a distance r from Q, we say that the electric potential at P is the scalar given by this formula:

    electric potential = k * (Q/r)

    My question is what if at some distance r from Q, the charge is very large. Wouldn't that then affect the electric potential. In other words how can this electric potential formula not take into account both charges and only takes into account Q which is the source charge?
  2. jcsd
  3. Aug 9, 2014 #2


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    It's just because elecitric potential is "energy per charge" and so is independent of the size of the charge (although the energy and force will vary at that same distance depending on the charge)

    It's very much like "gravitational potential" (not to be confused with "gravitational potential energy") which is independent of the mass of the object simply because it's defined as "energy per mass"
  4. Aug 9, 2014 #3
    So if it is independent of the size of the charge then how it is useful in solving problems to determine the change in electrical potential energy?
  5. Aug 9, 2014 #4
    Actually I am recalling from gravitation potential energy that you will get different values for PE depending on where you set PEgrav = 0 at. But if you try to find the change in gravitational potential energy that will be the same no matter where you set PEgrav = 0 at. Does the same idea apply here in the sense that if you find the change in gravitational potential energy it doesn't matter what you set as Q, the source charge or what q, the secondary charge is?
  6. Aug 9, 2014 #5


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    I'm probably not the best person to give a good example. But to me this seems like asking, "why is it useful to say 'g=9.8' for solving problems if that doesn't give the force on an object?"

    It's more for when you are interested in a certain position, irrespective of the charge (or mass) in it.
    (I'm sorry :smile: I'm sure there are much better explanations)

    This is only because when you find the "change" in gravitational potential energy, a final and initial position are implied. So you could pick any random PE=0 position, but it will always reduce to the "final position" becoming the PE=0 position
    (simply because the PE that they have in common (relative to your PE=0 position) is irrelevant, and "cancels out")
  7. Aug 9, 2014 #6
    The point charge is supposed to be a "test charge" that is very small compared to Q.


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