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nj1202

A semicircle of radius r is in the first and second quadrants, with the center of curvature at the origin. Positive charge +Q is distributed uniformly around the left half of the semicircle, and negative charge -Q is distributed uniformly around the right half of the semicircle. What is the magnitude of the net electric field at the origin produced by this distribution of charge? Express your answer in terms of the variables Q, r, and appropriate constants. Thanks in advance for any help in approaching this. Can't quite nab the hang of this one xD

  • one year ago
  • one year ago

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  1. nj1202
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    |dw:1346990452737:dw|

    • one year ago
  2. akash123
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    |dw:1346993325948:dw|

    • one year ago
  3. akash123
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    Take elementary charge -dq in the ist quadrant at angle theta...electric field E due to this electric charge at O: dE= -dq/ 4pi epsilon(0) r^2

    • one year ago
  4. akash123
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    and its dirction will be radially outward..

    • one year ago
  5. akash123
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    now break dE into tow components along x and y axis respectively...dE cos(theta) and dE sin(theta)

    • one year ago
  6. akash123
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    |dw:1346993841419:dw|

    • one year ago
  7. nj1202
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    So I actually was doing an easier problem earlier where this semicircle was all positive Q. And I found: \[E=\frac{ Q }{ 2\pi ^{2} \epsilon _{0}r ^{2}}\] Can I use this at all to help me in this?

    • one year ago
  8. akash123
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    can you show me the solution of this...I think answer will be something different...but how it'll help you in this?

    • one year ago
  9. Kainui
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    Let's simplify this by considering symmetry. Break up the electric field into x and y and see if any of them cancel. |dw:1346994485662:dw| Hey, what do you know, the y-coordinate cancels out and now you have a slightly easier problem to deal with.

    • one year ago
  10. Kainui
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    Not to mention you have the exact same contribution from the electric field from one and the other going to the right, so you really only need to solve for the electric field of one and multiply that by 2 to get the magnitude right.

    • one year ago
  11. Kainui
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    It's tempting to say that the electric field is related to 2 times the integral of cos theta from 0 to pi/2, but it's not I think lol.

    • one year ago
  12. nj1202
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    I typed in demitris's answer. It said that it was missing or had an incorrect numerical multiplier. Any idea what that could mean? he's probably very close to being right, the system is just being picky about what it has entered into it...

    • one year ago
  13. nj1202
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    What does your k = ? 1/4piEo ?

    • one year ago
  14. nj1202
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    Yea, okay. I just don't see how you went from the last part in the 2nd to last drawing to cancelling the 4s. Where is like Eo?

    • one year ago
  15. nj1202
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    But where did Eo disappear to in your last step?

    • one year ago
  16. nj1202
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    OH! cuz you had to double. so you wrote the constant, and then the numerator doubled. got it... but i still don't understand why Eo is missing....

    • one year ago
  17. nj1202
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    |dw:1346997418458:dw|

    • one year ago
  18. nj1202
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    Ok.. hopefully that does the trick and now it's right xD

    • one year ago
  19. nj1202
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    Just to let you know, what i wrote was right. it said correct online :) thank you so much for all the help !

    • one year ago
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