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Electric field and dipole

  1. Sep 9, 2016 #1
    Is this the correct equation to find the electric field at point p from a dipole?
    upload_2016-9-9_3-33-52.png
    upload_2016-9-9_3-34-2.png
     

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  3. Sep 9, 2016 #2

    BvU

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    Hello Novel, :welcome:

    You don't want to delete the template; it's very useful for you as well as for us. See the guidelines.

    Answer to your question: No. ##\ \vec E\ ## is a vector. What does your E describe, you think ?
     
  4. Sep 9, 2016 #3
    Thanks for your reply. I'll make sure to not delete the template next time. I think my E describes the magnitude of the electric field. After calculating the magnitude, I can decide if it's positive or negative. So in this case, E is negative, because E is going down, away from positive and toward negative. Is this right?
     
  5. Sep 9, 2016 #4

    BvU

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    I'll give you some leeway because you are new here (let's hope I don't get chastized for that).
    Also because I think you have a fair idea what you are doing, but you stumble because you are going too fast.

    Again, ##\vec E## is a vector. I've drawn the two contributions from the +Q and the -Q in the figure.
    There are no other contributions, so the field at P is the sum of these two. The vector sum, that is. Your job to do this vector addition. Andf yes, x/r appears in there (not x/R but x/r; I don't see or know of R in your post. Work accurately :smile:). And yes, it's downwards. Easy exercise, but a good vehicle to learn to work systematically.


    upload_2016-9-9_12-12-42.png
     
  6. Sep 9, 2016 #5
    I think I understand.

    E = kQ/r2 * cos(theta) because the y-components cancel out and we just want to get the x-component. I'll call the horizontal distance d (in a real problem it would be given or I could find it with trig), so cos(theta) = d/r. Therefore E = kQ/r2 * d/r = kQd/r3.
     
  7. Sep 9, 2016 #6
    I just realized my mistake. It's the x-components that cancel, not the y. So E = kQ/r2 * sin(theta) = kQ/r2 * x/r = kQx/r3 and then multiply by 2 because there are two charges acting on P in the same direction. So E = 2kQx/r3.
     
  8. Sep 9, 2016 #7

    BvU

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    Looks good to me. x and y are a bit confusing here because of the x's in the figure.
     
  9. Sep 9, 2016 #8
    Right, I realized that I should have chose a different variable name than x. Thank you for your help! I appreciate it!
     
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