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Want to know correct answer for electric field + potential question

  1. Jan 30, 2008 #1
    1. The problem statement, all variables and given/known data


    Three point charges are arranged on the y-axis as shown in the picture.

    The charges are:
    +q at (0,a)
    -q at (0,0)
    +q at (0,-a)

    Any other charge or material is infinitely far away.

    (a) Determine the point(s) on the x-axis where the electric potential due to this system of charges is zero.

    (b) Determine the x and y components of the electric field at a point P on the x-axis distance x from the origin.

    2. Relevant equations

    [tex]V = {k}_{e}\frac{q}{r}[/tex]

    [tex]E = {k}_{e}\frac{q}{r^2}[/tex]

    3. The attempt at a solution

    (a) The problem says the point(s) where potential is zero is must be on the x-axis so I know the y-coordinate is 0. I called the distance d, so my coordinate would be (d,0).

    I make this equation:

    [tex]0 = 2({k}_{e}\frac{q}{\sqrt{a^2+d^2}}) - {k}_{e}\frac{q}{d}

    And simplify/solve for d to get:

    [tex]d = \frac{\sqrt{3}a}{3}[/tex]

    (b) For this part I guess I just need to express the E vector in terms of x, q, ke, and a.

    Again I ignore any y components since the +q charges cancel in that direction.

    To get the magnitude of the field vector from the (+) charges I plugged in the distance of P (using a right triangle with sides x and a) from the charge for r and doubled:

    [tex]E = 2({k}_{e}\frac{q}{(\sqrt{a^2+x^2})^2})[/tex]

    and to get the x component multiplied that by:

    [tex]{E}_{x} = \cos\tan^{-1}\frac{a}{x}[/tex]

    Then I find the net to be (including E from the (-) charge):

    [tex]{E}_{x net} = 2(\cos\tan^{-1}\frac{a}{x})({k}_{e}\frac{q}{a^2+x^2}) - {k}_{e}\frac{q}_{x^2}[/tex]

    [tex]{E}_{x net} = \frac{2{k}_{e}q(\cos\tan^{-1}\frac{a}{x})(x^2) - {k}_{e}q(a^2+x^2)}{(a^2+x^2)x^2}[/tex]

    I feel ok about (a) but I might be wrong. As for (b), I have no idea if I'm right or not but I did try it multiple ways to get an equivalent solution. It just looks too complicated though. Please let me know if I have done these right.
    Last edited: Jan 30, 2008
  2. jcsd
  3. Jan 31, 2008 #2

    Shooting Star

    User Avatar
    Homework Helper

    3a) d should have two values.

    3b) What happened to the field due to the charge (-q)?
  4. Jan 31, 2008 #3
    Thanks for the reply!

    Ah yeah that makes sense for (a). It should be plus or minus then.

    For (b), would the E from -q be [itex]-{k}_{e}\frac{q}_{x^2}[/itex]?
  5. Feb 1, 2008 #4

    Shooting Star

    User Avatar
    Homework Helper

    Write cos theta = x/sqrt(x^2+a^2).

    That is correct now.
  6. Mar 3, 2008 #5
    can't you just replace [tex](\cos\tan^{-1}\frac{a}{x})[/tex] with [tex]\frac{x}{\sqrt{(x^2+a^2)}}[/tex]

    ...or is that what shootingstar was saying
    Last edited: Mar 3, 2008
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