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Homework Help: Force from two point charges on a third one

  1. May 20, 2014 #1
    1. The problem statement, all variables and given/known data

    We have three point charges [itex] Q_1, Q_2, Q_3 [/itex], forming together an isosceles triangle. Two sides are of equal length [itex] r [/itex].
    The point charge [itex]Q_1[/itex] is located at [itex] (0,0)[/itex],
    the point charge [itex]Q_2[/itex] is located at [itex] (x_2,0)[/itex]
    and The point charge [itex]Q_3[/itex] is located at [itex] (\frac{x_2}{2},\sqrt{r^2 - ({\frac{x_2}{2}})^2})[/itex].

    Calculate the resulting force on [itex]Q_3[/itex] for the cases [itex] Q_2 = Q_1 [/itex] and [itex] Q_2 = - Q_1. [/itex]

    2. Relevant equations

    I think two formulas are relevant. Coulombs law and the superposition principle.

    3. The attempt at a solution

    From the Superposition principle and Coulombs law we have:

    [tex] \vec{F}_3 = \frac{1}{4 \pi \epsilon_0} Q_3 Q_1 \frac{\vec{x}_3 - \vec{x}_1}{|\vec{x}_3 - \vec{x}_1|^3} + \frac{1}{4 \pi \epsilon_0} Q_3 Q_2 \frac{\vec{x}_3 - \vec{x}_2}{|\vec{x}_3 - \vec{x}_2|^3},[/tex]
    where [itex]\vec{F}_3[/itex] is the resulting force on [itex]Q_3[/itex].

    Next, I calculated [itex] \vec{x}_3 - \vec{x}_1 = (\frac{x_2}{2},\sqrt{r^2 - ({\frac{x_2}{2}})^2}) [/itex]

    and [itex] \vec{x}_3 - \vec{x}_2 = (-(\frac{x_2}{2}),\sqrt{r^2 - ({\frac{x_2}{2}})^2}). [/itex]

    Then I calculated the norm of both to get [tex] |\vec{x}_3 - \vec{x}_1| = |\vec{x}_3 - \vec{x}_2| = r. [/tex]

    So I get for [itex] \vec{F}_3 [/itex],

    [tex] \vec{F}_3 = \frac{1}{4 \pi \epsilon_0} \frac{Q_3 Q_1}{r^3} (\frac{x_2}{2},\sqrt{r^2 - ({\frac{x_2}{2}})^2}) + \frac{1}{4 \pi \epsilon_0} \frac{Q_3 Q_2}{r^3} (-\frac{x_2}{2},\sqrt{r^2 - ({\frac{x_2}{2}})^2})[/tex]

    Now for [itex] Q_2 = Q_1 [/itex]:

    [tex] \vec{F}_3 = \frac{1}{4 \pi \epsilon_0} \frac{Q_3 Q_1}{r^3}( (\frac{x_2}{2},\sqrt{r^2 - ({\frac{x_2}{2}})^2}) + (-\frac{x_2}{2},\sqrt{r^2 - ({\frac{x_2}{2}})^2}))[/tex]

    [tex] \vec{F}_3 = \frac{1}{4 \pi \epsilon_0} \frac{Q_3 Q_1}{r^3}( 0,2 \sqrt{r^2 - ({\frac{x_2}{2}})^2}).[/tex]

    And for the case [itex] Q_2 = - Q_1 [/itex]:
    [tex] \vec{F}_3 = \frac{1}{4 \pi \epsilon_0} \frac{Q_3 Q_1}{r^3} (x_2,0).[/tex]

    Is this correct? What do you think?
    Last edited: May 20, 2014
  2. jcsd
  3. May 20, 2014 #2


    Staff: Mentor

    Can you edit your post and remove all the extra blank homework template stuff?

    It will be easier on the eyes.
  4. May 20, 2014 #3
    Ok! Done!
  5. May 20, 2014 #4


    Staff: Mentor

    it looks right. Have you plugged some numbers in and sketched the vectors on paper to see if the result looks reasonable?

    When the two charges are the same then the third charge will either move away or be attracted to the other two. In either case, the force will be along the center line through the isoseles triangle.

    When the two charges are different then the third charge will move on a line going thru the third charge and parallel to the line going thru the two other charges initially. The resultant force should be parallel to these lines.
  6. May 21, 2014 #5
    Thanks jedishrfu! I'll plug in some numbers and draw the vector to reproduce your argument!
    Thanks again!
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