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Homework Help: Calculating the electric field, multiple point charges

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data

    What are the strength and direction of the electric field at the position indicated by the dot?


    Part A:
    Give your answer in terms of the horizontal and vertical components, separated by commas. Take the positive directions to be up and to the right.

    Part B:
    Specify the strength of the electric field.

    2. Relevant equations


    3. The attempt at a solution





    E1=112500 j
    E2=56250 i

    E3_i=45000*(0.04/0.0447)=40249 i
    E3_j=45000*(0.02/0.0447)=20134 j

    Right is positive, therefore:
    E2= -56250 i
    E3_i= -40249 i

    ETotal_i= -56250 + -40249 = -96499 i
    ETotal_j= 20134 + 112500 = 133534 j

    This is coming out as 'incorrect', and therefore I cannot proceed to part 2

    Where am I going wrong?

    Last edited: Oct 7, 2007
  2. jcsd
  3. Oct 7, 2007 #2
    You need to redo E3.
  4. Oct 7, 2007 #3
    typo, sorry, E3 is actually 45000, unless you mean that that is also incorrect
    Last edited: Oct 7, 2007
  5. Oct 7, 2007 #4
    Oh, okay. 45000 is correct.

    I just calculated the x-comp. of E3 and it comes out to 40,268.45638, while your answer is off by almost 20 N/C. I think it's a calculation error somewhere.
  6. Oct 7, 2007 #5
    yea idk for some reason my calculator is giving me the 40249.22359..., anyways, is my y-component correct? i have very few attempts left

  7. Oct 7, 2007 #6
  8. Oct 7, 2007 #7
  9. Oct 7, 2007 #8
    ....still isnt working :(
  10. Dec 13, 2007 #9
    The answer is E(0,0.2) = (-964.15,-922.9) where E(0,0.2) = 1334.7
    Last edited: Dec 13, 2007
  11. Dec 13, 2007 #10
    please correct the value of r3, then everything should fall into place.
  12. Jan 25, 2008 #11
    I got (-96392,-92272)

    your J value is incorrect. your e1 is incorrect. using the correct sign convention, since e1 is a negative charge, it pulls the dot towards it. not away, therefore e1 should be negative since up and right is postive
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