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Coulomb's Law problem with three charged points

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data
    TRHZ2.jpg

    2. Relevant equations

    F=Keq1q2/r2

    3. The attempt at a solution

    Using Coulomb's Law, here's what I've worked out:

    FAB=0.593N
    FAC=0.309N

    Fx=0.309 - 0.593cos60°= 0.0125
    Fy=0.593sin60° =0.514

    √(0.01252 + 0.5142) = 0.514 N/C

    and θ = tan-1(0.514/0.0125) = 88.6°

    I'm not sure why webassign is telling me that my answer is wrong. I worked out a problem almost exactly like this with slightly different numbers and got it right...
     
  2. jcsd
  3. Sep 15, 2012 #2
    You're finding the force on particle A, but the problem asks for the E-field at the position of particle A. Do you know the equation for the E-field of a point charge?
     
  4. Sep 15, 2012 #3
    I thought it was the same formula? Shoot. Idk.

    I'll review the problem and post here again if I have trouble. I guess I was just looking for the wrong thing. Thanks for pointing that out.
     
  5. Sep 15, 2012 #4
    It's a similar formula, but the E-field has units of N/C (while the equation you used gives force in N).

    Hope that helps. If you have more trouble with your problem don't hesitate to ask.
     
  6. Sep 16, 2012 #5
    So would the E-field simply be:

    E=ke(2.2x10-6)/(0.52) = 7.91 x 104 N/C?
     
  7. Sep 16, 2012 #6

    SammyS

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    Webassign is right to tell you it's wrong.

    However, Webassign (or whoever wrote the problem) should have asked for the electric field at the position of the 2.2μC due to the two other charges.

    Back to your issue ...

    What is the direction of the electric field due to charge B?

    What is the direction of the electric field due to charge C?
     
  8. Sep 17, 2012 #7
    Ok, I gave it a try but I think I'm still wrong.

    EA = 7.91 x 104 N/C
    EB = 2.70 x 105 N/C
    EC = 1.40 x 105 N/C


    EAX = (2.70 x 105) + (7.91 x 104)cos 60 = 3.10 x 105

    EAY = (2.70 x 105) + (7.91 x 104)sin 60 = 3.39 x 105

    ECX = (2.70 x 105) - (1.40 x 105)cos 60 = 2.00 x 105

    ECY = (2.70 x 105) - (1.40 x 105)sin 60 = 1.49 x 105

    Ex = EAX - ECX = 3.10 x 105 - 2.00 x 105 = 1.10 x 105

    EY = EAY - ECY = 1.90 x 105

    So, √[(1.10 x 105)2 + (1.90 x 105)2)]

    = 2.20 x 105 N/C

    And [itex]\theta[/itex] = tan-1(EY/EX) = 60°

    Phew. I hope that's at least in the right ballpark!? I got really confused about when I should be adding and subtracting some of those vectors....
     
  9. Sep 17, 2012 #8
    Crap. I think I already made a mistake...I was supposed to ADD ECX...

    Doing that gave me a final answer of 1.92 x 105 N/C
     
  10. Sep 18, 2012 #9
    Well I couldn't figure it out in time. This was the only HW problem I missed luckily. Apparently the answer is 2.34 x 10 5

    I wasn't too far off with my first answer...if anyone could help me with the solution I'd be really thankful.
     
  11. Sep 18, 2012 #10

    SammyS

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    Charge B produces an E field of, [itex]\displaystyle \vec{E}_B=k_e\frac{7.5\times10^{-6}}{(0.500)^2}(\cos(120^\circ)\hat{i}+\sin( 120^\circ)\hat{j})=k_e\frac{7.5\times10^{-6}}{(0.500)^2}\left(-\frac{1}{2}\hat{i}-\frac{\sqrt{3}}{2}\hat{j}\right)[/itex]

    Charge C produces an E field of, [itex]\displaystyle \vec{E}_C=k_e\frac{3.9\times10^{-6}}{(0.500)^2}\hat{i}[/itex]

    Therefore, [itex]\displaystyle \vec{E}=\frac{k_e\cdot 10^{-6}}{(0.500)^2}\left((3.9-3.75)\hat{i}
    -3.75\sqrt{3}\hat{j}\right)
    [/itex]
     
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