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Moving point charges - help debugging

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data

    See attachment

    2. Relevant equations



    3. The attempt at a solution

    I was successful with finding a solution. But, only because I tried the practice version of that (where you can see the answers) and noticed the equation I derived gave angles that, when summed with the correct answer, yielded 180*.

    Can someone tell me why?

    START:

    Since Bead 1 is stationary, we know its electric field will be in the direction of the positive x axis.

    [itex]E_{1} = \frac{1}{4\pi\epsilon_{0}} \frac{q_{1}}{r^{2}} \hat{i}[/itex]

    While Bear 2 can be moved, thus its field will be a function of theta.

    [itex]E_{1} = \frac{1}{4\pi\epsilon_{0}} \frac{q_{2}}{r^{2}}cos(\theta) \hat{i} + \frac{1}{4\pi\epsilon_{0}} \frac{q_{2}}{r^{2}}sin(\theta) \hat{j}[/itex]

    Thus, we add these two vector equations together for net E, and take the magnitude. After a bit of algebra we get:

    [itex]E^{2}(4\pi\epsilon_{0})^{2} = q^{2}_{1} + 2q_{1}q_{2}cos(\theta) + q^{2}_{2}cos^{2}(\theta) + q^{2}_{2}sin^{2}(\theta)[/itex]

    By factoring out the q2 squared and noting the trigonometric identity,

    [itex]E^{2}(4\pi\epsilon_{0})^{2} = q^{2}_{1} + 2q_{1}q_{2}cos(\theta) + q^{2}_{2}[/itex]

    [itex]E^{2}(4\pi\epsilon_{0})^{2} - q^{2}_{1}- q^{2}_{2}= 2q_{1}q_{2}cos(\theta)[/itex]


    [itex]\frac{E^{2}(4\pi\epsilon_{0})^{2} - q^{2}_{1}- q^{2}_{2}}{2q_{1}q_{2}}= cos(\theta)[/itex]

    For a final formula of:

    [itex]arccos(\frac{E^{2}(4\pi\epsilon_{0})^{2} - q^{2}_{1}- q^{2}_{2}}{2q_{1}q_{2}}) = \theta[/itex]


    However, the value of the number within the arccos function is the negative of what it should be.

    This gives me the answer correctly:

    [itex]arccos(- \frac{E^{2}(4\pi\epsilon_{0})^{2} - q^{2}_{1}- q^{2}_{2}}{2q_{1}q_{2}}) = \theta[/itex]

    Why?
     

    Attached Files:

  2. jcsd
  3. Aug 25, 2012 #2

    TSny

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    Homework Helper
    Gold Member

    Consider the direction of the electric field of bead 2 and decide on the signs of its x and y components.
     
  4. Aug 25, 2012 #3
    I think I understand you.

    While cos(45) is positive, at an angle 45 B's electric field is pointing down and towards the left.

    Right?
     
  5. Aug 25, 2012 #4

    TSny

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    Homework Helper
    Gold Member

    Yes :smile:
     
  6. Aug 25, 2012 #5
    Thank you very much.
     
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