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Electric Potential of two charges

  1. Nov 2, 2007 #1
    1. The problem statement, all variables and given/known data
    Two charges lie on the x axis, +3q at the origin, and -2q at x=5.0m. The point on the x axis where the electric potential has a zero value (when the value at infinity is also zero) is

    2. Relevant equations

    V = kQ/r
    U = qV
    U = q[tex]_{1}[/tex]kq[tex]_{2}[/tex]/r

    3. The attempt at a solution

    Not sure at all.

    If I substitute V = 0 in those equations, r will turn out to be infinity.

    But the answers are multiple choice:

    a. 1.0 m.
    b. 2.0 m.
    c. 2.5 m.
    d. 3.0 m.
    e. 4.0 m.
  2. jcsd
  3. Nov 2, 2007 #2
    Hi there

    I'm not sure if i'm right.

    I think the easiest way is to use your gut feel. That is it must be nearer to -2q than 3q since further away from 3q, the effect of the 3q charge would be less.

    Your approach looks right to me. But you must keep in mind that when you sub V = 0 in, you will need to have a more complex equation. That is, you have two distances. One distance, x the distance from 3q to the point of equilibrium and (5-x) the distance from -2q to the point of equilibrium.

    I think that will solve your problem.
  4. Nov 2, 2007 #3
    Nope, tried that just now, wrong answer, but I'm sure I did it wrong, cause I'm not really sure what you meant
  5. Nov 2, 2007 #4
    Did you mean setting:

    kQ/x = kQ/5-x

    And then solving for x?
  6. Nov 2, 2007 #5
    yes, that is what i meant.
  7. Nov 2, 2007 #6
    Hi. Actually, I think it's supposed to be (x+5).
    (this is wrong)

    No, it's (x-5) same idea as transformation on graphs I think. I'm sorry.
    Last edited: Nov 2, 2007
  8. Nov 2, 2007 #7
    For that matter, I don't think you can let V = 0 because V is a scalar. I think we should set E (electric field strength) be zero and solve for x. However, when I do this I do not get the answer.
  9. Nov 2, 2007 #8
    It's ok, I'll just take an educated guess
  10. Nov 2, 2007 #9
    This was a multiple choice - there are several ways to eliminate the ridiculous/unlikely, and clearly several of you have felt unsure or confused. It comes from not understanding the difference between the electric field due to a point charge, and the electric potential energy.

    Electric fields reduce as inverse square law of the distance from the charge, and they determine the mechanical force to another charge (with its own field). electric potential is about the work done on moving a charge over a distance against the force to another charge. Its how we define the volt, where we deliberately use a unit charge. It scales as the straightforward inverse as distance from charge.

    Start with your best instincts.
    3q at x=0 across to -2q at x=5. That lets out x=2.5 for a start. Halfway between the charges would only make sense if they had the same magnitude.

    The effect of a charge is going to be proportional to the charge, and inversely proportional to the distance. Consider both the 1m and 4m answers. x=1m is 1m close to a 3q charge, with a -2q charge all of 4m away, and x=4m is 1m close to a -2q charge with +3q also 4m away.

    Since |3q|/1m is not equal to |2q|/1m, it leaves only x=3, since the only way to balance the unequal potentials associated with these charges, is to be at distances in the same ratio. One of the choices is exactly that!
    Last edited: Nov 2, 2007
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