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Simple electrostatics

  1. Feb 12, 2005 #1
    I've got the following problem:
    1) We've got a charge -Q at location at the origin and two charges of the same magnitude but opposite sign at [tex](a,0,0)[/tex] and [tex](0,a,0)[/tex]. Determine the total force on the charge at the origin.

    2) We've got the same charge -Q at the origin but now a charge of 2Q at [tex](a,a,0)[/tex]. Again, determine the force on the charge at the origin.
    Then compare your answer to the previous question

    The first thing I did was to calculate the seperate forces and then add them together.

    [tex]\vec{F}_1 = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{a^2} \hat{x}[/tex]
    [tex]\vec{F}_2 = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{a^2} \hat{y}[/tex]

    So the total force is given by
    [tex]\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{a^2} (\hat{x}+\hat{y})[/tex]

    Now for the second question the distance between the two charges is given by [tex]\sqrt{a^2+a^2}[/tex], so the force is given by
    [tex]\vec{F} = \frac{2Q^2}{4\pi\epsilon_0}\frac{a\hat{x} + a\hat{y}}{(a^2+a^2)^{\frac{3}{2}}} = \frac{2Q^2}{4\pi\epsilon_0}\frac{a\hat{x} + a\hat{y}}{2\sqrt{2}a^3} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{a^2}\frac{1}{\sqrt{2}}(\hat{x} + \hat{y})[/tex]

    Now I would like to ask you if these calculations are really correct? :smile:
     
  2. jcsd
  3. Feb 12, 2005 #2

    Doc Al

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    Staff: Mentor

    Perfectly correct. You may want to write your answer as a magnitude times a unit vector:
    [tex]\vec{F} = \sqrt{2} \frac{1}{4\pi\epsilon_0} \frac{Q^2}{a^2} [\frac{1}{\sqrt{2}}(\hat{x}+\hat{y})][/tex]

    Again, perfectly correct.
     
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