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Calculating Electric Field

  1. Feb 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Two 1.00µC charged particles are located on the x axis. One is at x = 1.00 m, and the other is at x = -1.00 m.
    Determine the electric field on the y axis at y = 0.400 m

    2. Relevant equations
    Electric Field = ke * q / r^2

    3. The attempt at a solution

    I know the x-components cancel out, so we're only left with the y-component.
    E= 8.99*10^9 * (1*10^-6 C / (square root(1^2 + .400^2)) and then you have to take the y-component into account, so you multiply that by sin(.400/(square root(1^2 + .400^2))... and then multiply everything by 2 because there are two different charged particles. I'm getting an answer of 5625 N/C, but apparently the answer is 5760 N/C.
     
  2. jcsd
  3. Feb 17, 2016 #2

    SammyS

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    According to the following, you only divided by r, not r2 .

    E= 8.99*10^9 * (1*10^-6 C / (square root(1^2 + .400^2))
     
  4. Feb 17, 2016 #3
    When I put it in my calculator, I did square it... oops, just forgot to type it here. Still getting 5625.
     
  5. Feb 17, 2016 #4

    SammyS

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    Are you rounding off any intermediate results?
     
  6. Feb 17, 2016 #5
    Nope, I put it in exactly how I did here without rounding off the calculation of the hypotenuse.
     
  7. Feb 17, 2016 #6

    SammyS

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    Maybe this is the problem. The following

    .400/(square root(1^2 + .400^2))

    is the sine of the angle. Don't take the sine of this. That would be sin(sin(θ)).
     
  8. Feb 17, 2016 #7
    What do you mean? I thought that since sine is OPP/HYP and the opposite would be 0.400 and the hypotenuse would use Pythagorean theorem and it would be square root(1^2 + .400^2, so it would be sin(.400/(square root(1^2 + .400^2))... right?
     
  9. Feb 17, 2016 #8

    SammyS

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    Yes. That, 0.400/square root(1^2 + .400^2), IS the sine. Don't take the sine of the sine.
     
  10. Feb 17, 2016 #9
    Ohhhh wow, that was it! Thank you so much! I didn't even realize.
     
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