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Help with Electric Forces Problem and Equilateral Triangles

  1. Feb 7, 2016 #1
    1. The problem statement, all variables and given/known data
    The point charges in the figure have the following values: q1=+2.1μC, q2=+6.3μC, q3=−0.89μC. Suppose that the magnitude of the net electrostatic force exerted on the point charge q2 in the figure is 0.57 N .
    Find the distance d and the direction (angle) of the net force.
    Walker.19.25.jpg
    2. Relevant equations
    (I think) F=k|q||q|/d^2
    and pythagorean theorem

    3. The attempt at a solution
    I really think I overcomplicated myself in the beginning (I separated each F12 and F23 into its x and y components), but my most recent solution goes:
    F12 = k|q1||q2|/d^2
    F23 = k|q2||q3|/d^2

    and then attempting to do (0.57)^2 = (k|q1||q2|/d^2)^2+(k|q2||q3|/d^2)^2
    and solving for d. I know the right answer is 0.43 m but I truly don't know how that is (neither of my two methods have worked). I've been doing this for two hours (to my embarrassment) so any input would be very much appreciated.

    Thanks in advance.
    (Also, is responding to these three questions what using the template means?)
     
  2. jcsd
  3. Feb 7, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes, that's the way to approach it. If you show us some details of what you did, maybe we can spot the error.
    This won't work because the forces F21 and F23 are not perpendicular to one another.

    Using the template just means filling in all three parts of
    1. The problem statement, all variables and given/known data
    2. Relevant equations
    3. The attempt at a solution

    You did that very nicely!:smile:
     
  4. Feb 7, 2016 #3
    Awesome, thanks!

    Ok, so this is what I did when I decomposed the vectors the first time:

    F12x= cos60F12
    F12y= sin60F12

    and

    F23x= F23
    F23y= 0

    giving me:

    FNetX = F23X-F12X (because they're going opposite I used the negative sign)
    FNetY = F12Y

    And then I did the pythagorean theorem for all of that:
    (F23X-F12X)2+(F12Y)2=FNet2

    [(k|q2||q3|/d^2 - cos60(k|q1||q2|/d^2)]2 + [sin60(k|q1||q2|/d^2)]2 = FNet2

    :eek:...and it works now...
    I wish I hadn't erased my first attempt (then I would've know what went wrong), but surely it was the part with the FNetX because I don't remember getting a negative answer to that before...wow.

    Yes *facepalm*, it seems silly to have given up on my previous plan for that one!

    Thank you so much TSny, and sorry about that! :oops:
     
  5. Feb 7, 2016 #4

    TSny

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    Homework Helper
    Gold Member

    Good. (I'm assuming you're taking the x axis to be horizontal and the y axis vertical.)

    This doesn't look right. F23 does not point along the x axis.

    Just to make sure we're together, can you describe in words the directions of F12 and F23 acting on q2?
     
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