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Calculating electric potential

  1. Jan 22, 2008 #1
    1. The problem statement, all variables and given/known data


    Each side of the triangle has a given length "d." Calculate the electric potential at point P.

    2. Relevant equations

    [tex] V(\vec{r}) = \frac{kq}{r} [/tex]

    I believe this is for the elec. potential at some distance r for a point charge q, but I'm not completely sure.

    3. The attempt at a solution

    My main problem is whether or not to take the sign of each charge into account when I'm summing the voltage. For instance, when considering the charges horizontal to the point P, +3q and - 3q, if I take the signs into account, I end up with zero voltage. I've read that the idea of superposition applies, but the voltage is a scalar, and needs to be handled differently.

    A push in the right direction would be much appreciated.
  2. jcsd
  3. Jan 22, 2008 #2


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    Homework Helper

    Calculate the electric potential due to each charge taking into account its sign.
    and add them.
    when considering the charges horizontal to the point P, +3q and - 3q, if I take the signs into account, I end up with zero voltage It is right.
  4. Jan 23, 2008 #3
    Ok, lets just say that I had the +3q and -3q charges, and the point P midway between them. If I say that the electric potential at P is +3e + (-3q) = 0, then I have 0 potential at that point. This seems strange to me. If I placed a positive test charge at point P, it would move away from +3q and move toward -3q, which infers that it has non-zero potential enery, and thus non-zero electric potential.
  5. Jan 23, 2008 #4
    You're correct in being suspicious of that hotcommodity

    Remember one way to think of electric field is force per unit charge

    So if you put a unit charge there and there's a force(like you said), there's clearly electric potential there
  6. Jan 23, 2008 #5
    So how do I add the voltages? I assume that charge plays role in determining whether there's positive or negative voltage at any one point, but I'm not sure how to apply that concept. Additionally I'm wondering if any angles need to be taken into account when adding the voltages.
  7. Jan 23, 2008 #6
    For the question you asked, with just the two charges in a line, then no, no need to worry about angles

    Remember what r is, it's the distance between the source(the charge)and the point(in this case point P), so assuming everything is lined up on the x axis

    Let's say the -3q is on the left at x=-3 and the positive 3q is at x=3 and point P is at the origin

    You'd have the electric potential at P because of the left charge be kq/r, q=-3q and r = -3....

    See where this is going?
  8. Jan 23, 2008 #7
    So given those values of r, I'd have an electric potential of 2q at P?
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