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Four Electric Dipoles = What level of electric flux?

  1. Oct 1, 2016 #1
    1. The problem statement, all variables and given/known data
    A circle, centered on the origin, has a radius of 1 mm. At each "pole" (1,0), (0,-1), (-1,0), (0,1) is an electric dipole. The positive charge of +10 microCoulombs is inside the circle, the negative charge of -10 microCoulombs is just outside the circle.

    What is the electric flux for the entire circle?

    2. Relevant equations
    Flux = q(encl)/e0

    3. The attempt at a solution
    I think the way to solve this is to simply multiple the + charge by four, as that's the enclosed charge, then divide by epsilon naught. Is this the correct way to go about this? Am I missing something? I thought this would be more difficult than that simple of a solution.
  2. jcsd
  3. Oct 1, 2016 #2


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    Are all these dipoles lying in the same plane as the circle? Assuming that they are, what is the orientation of the dipoles relative to each other and relative to the plane of the circle? I assume that the required flux is through the area of the circle in which case you have to use the definition of flux and not Gauss's law. You need a closed surface to enclose a charge and here there isn't one.

    On edit: I have a feeling this is not a well-posed question.
  4. Oct 1, 2016 #3
    This is a diagram that's like the one given. The space between the + and - charges is 1 mm. They're all in the same plane. Like I mentioned in the OP, the positive four charges are on the inside of the circle with the corresponding - charges on the outside. The question was, what's the flux through the circle.

    I hope this helps!

  5. Oct 1, 2016 #4


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    This helps. I thought that "positive" charges meant additional point charges separate from the dipoles. My mistake.

    Do you know what the E-field due to a dipole looks like? Draw some electric filed lines and see what you can say about the electric flux, specifically about ## \vec{E} \cdot \hat{n} ## right on the surface of the circle.
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