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Potential Due to a Collection Charges

  1. May 16, 2008 #1
    My text reads for Potential Due to a Collection Charges:

    1/(4πє) ∑ q/r

    Lets say you want to calculate the potential between two charges. Do you take the magnitude of the distance (r) or do you account for their direction of the charges with respect to the field point.

    Based on the examples I have seen, i seems like i should use the magnitude. However, it seems more logical to account for direction.
  2. jcsd
  3. May 16, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    potential is a scalar

    The potential due to two charges is exactly that given by the formula you provided, thus:
    [tex]\frac{1}{4\pi\epsilon_0} (\frac{q_1}{r_1} + \frac{q_2}{r_2})[/tex]

    r_1 & r_2 are the distances from each charge. (Direction is not relevant, only distance from each charge.)
  4. May 16, 2008 #3
    Woops, I didn't read the forum rules. I appreciate your help Doc Al. This won't happen again.
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