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anonymous

  • 4 years ago

just need this answer real quick...timed test..got 13 minutes left: A +48.26 nC point charge is placed on the x axis at 3.007 m, and a -21.49 nC point charge is placed on the y axis at y= -6.779 m. What is the direction of the electric field at the origin?

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  1. anonymous
    • 4 years ago
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    counterclockwise from the positive x axis

  2. anonymous
    • 4 years ago
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    plz tell me u can do this?

  3. ash2326
    • 4 years ago
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    yeah let me try

  4. ash2326
    • 4 years ago
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    |dw:1328301704314:dw|

  5. anonymous
    • 4 years ago
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    i got 8 minutes to submit answer

  6. anonymous
    • 4 years ago
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    i hate timed tests

  7. anonymous
    • 4 years ago
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    Just think about it yourself. First hint: You can regard the resulting electrical field as the sum of the two electrical fields of each charge.

  8. anonymous
    • 4 years ago
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    chris i don't have the time...i got 7 minutes

  9. anonymous
    • 4 years ago
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    Enough time to think ...

  10. anonymous
    • 4 years ago
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    afterwards i'll worry about how's and why's...right now i just need the correct answer

  11. anonymous
    • 4 years ago
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    26.77 is net electric charge

  12. ash2326
    • 4 years ago
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    Electric field at origin due to positive charge=9*10^9*48.26*10^-9/(3.007)^2( -i) = 48.0355 (-i) (toward left ) Electric field at origin due to negative charge=9*10^9*(-21.49)*10^-9/(6.779)^2 (j) =4.208(-j) towards downward direction I think clockwise from x direction

  13. anonymous
    • 4 years ago
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    Second hint: Use coulomb's law to calculate the contribution of each charge at the origin.

  14. anonymous
    • 4 years ago
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    plz for the love of god just gimme the answer and i'll work through it after i submit it

  15. anonymous
    • 4 years ago
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    plz

  16. ash2326
    • 4 years ago
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    got it?

  17. anonymous
    • 4 years ago
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    I'm not going to support you to cheat, sorry.

  18. anonymous
    • 4 years ago
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    then leave

  19. anonymous
    • 4 years ago
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    ashe no i don't

  20. ash2326
    • 4 years ago
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    are there options

  21. anonymous
    • 4 years ago
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    Damn, just add these numbers.

  22. anonymous
    • 4 years ago
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    what numbers?

  23. anonymous
    • 4 years ago
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    i got 2 minutes left?

  24. anonymous
    • 4 years ago
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    u wnana throw an airball up?

  25. ash2326
    • 4 years ago
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    net electric field= \[\sqrt{{48.0355}^2+{4.028}^2}\]=48.204 Volt/meter in clockwise direction from =ve x-axis

  26. ash2326
    • 4 years ago
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    it'd be +ve x -axis

  27. anonymous
    • 4 years ago
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    i submitted 128...i think it's 132 though

  28. anonymous
    • 4 years ago
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    thx for your help sir

  29. ash2326
    • 4 years ago
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    but from where you got 128

  30. anonymous
    • 4 years ago
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    i appreciate the effort

  31. anonymous
    • 4 years ago
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    i added the two numbers u had earlier....as per what chris said

  32. anonymous
    • 4 years ago
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    then just subtracted taht from 180

  33. anonymous
    • 4 years ago
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    to get the angle counterclockwise

  34. ash2326
    • 4 years ago
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    angle is -175 degrees

  35. ash2326
    • 4 years ago
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    that was magnitude

  36. anonymous
    • 4 years ago
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    If ash's numbers are correct it's just \[\vec E = ({-48.0},{-4.2} )^T \frac{\textrm{N}}{\textrm{C}}\], i.e. 180.5° (counter-clockwise).

  37. anonymous
    • 4 years ago
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    yea thx....coulda used that 15 minutes ago

  38. anonymous
    • 4 years ago
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    good lookin out

  39. Xishem
    • 4 years ago
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    In the future you might consider: a.) Not wait until the last minute to do a test like this. b.) Being respectful to those who helped you. They aren't required to do so.

  40. anonymous
    • 4 years ago
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    to xishem....i did not wait until the last minute to do this test. It's a timed 20 minute 1 question test. I'm very respectful of those who help me. I help many many people on here in the math section. I was very respectful and appreciative of ash for taking the time to help me. Chris was not helping, therefore at the time, was not appreciative

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