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Dipole problem help

  1. Feb 18, 2008 #1
    !!!dipole!!!

    [​IMG]

    Okay. So I am trying to do this kind of the old fashioned way. I broke the E's down into x and y component and am adding them together component-wise.

    I have found that the x components cancel out. Now I am trying to widdle down the form I have for the y components. As of now I have that:

    [tex]E_y=k[\frac{-|q_1|-|q_2|}{[(d/2)^2+r^2]^{1/2}}*\frac{d/2}{[(d/2)^2+r^2]^{1/2}}[/tex]


    The text has it in the form: [tex]kqd/r^3[/tex] which I am having a hard time getting.

    How do I get rid of the negative sign? AND how do I make use of the fact r>>d??
     
  2. jcsd
  3. Feb 18, 2008 #2
    Anyone??
     
  4. Feb 18, 2008 #3

    hage567

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    Homework Helper

    q1 and q2 are equal.

    In your first term you have the denominator to the 1/2. That shouldn't be there. Check your work.

    I'm not sure what to say about the negative sign because I'm not sure how you came up with it.

    If r>>d then you can ignore your (d/2)^2 term in the denominator because it will be negligible compared to r^2.
     
    Last edited: Feb 18, 2008
  5. Feb 18, 2008 #4
    I got the negative sign when I drew the free body diagram of my test charge. Since q_o is +, both fields are in the negative y direction.

    Or should I just be considering their magnitude of 2q in this case and forget which direction they point?

    It is to the 1/2 because it is the distance from each q to p [itex]c^2=b^2+a^2[/itex]
    Thanks!
     
    Last edited: Feb 18, 2008
  6. Feb 18, 2008 #5

    hage567

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    OK, that's just a matter of convention then. You could have made it positive if you wanted.

    Since the y components are in the same direction (and equal magnitude), you can just say that E = 2*Ey (where Ey is the y component from one of the charges).

    The general equation is [tex]E = \frac{kq}{x^2}[/tex] right?

    So you found the length of the hypotenuse using Pythagorus, so [tex]x^2 = (\frac{d}{2})^2 + r^2 [/tex]

    So there is no need to take the square root to get just x, since you need x^2 in your equation anyway. Does that make sense? Just the first term in your denominator is wrong, the second one is right.
     
  7. Feb 18, 2008 #6
    Oh crap-ass! I do crap like this all the time. . . Thanks hage!!
     
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