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Electric field of a loop with one positive halve and one negative halve

  1. Aug 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Given a circular wire with radius R. Choose the origin in the center of the circle, the z-axis perpendicular to the circle. One halve of the circle contains a positive line charge [itex]\lambda[/itex], the other halve a negative line charge of the same magnitude.
    (a) Sketch the electric field lines.
    (b) Calculate the electric field in a point on the positive z-axis and argument that it contains only a component in the z direction.
    (c) What is the direction of the electric field in the x-direction, and with what power of r does this field fall off?
    I hope I have correctly translated this from Dutch. ;)

    2. Relevant equations
    [itex]\vec{E} = \frac{1}{4 \pi \epsilon_0}\int \frac{\hat{r}}{r^2}\lambda dl'[/itex]


    3. The attempt at a solution
    (a) I have no idea how to do this. I thought because of the opposing charges the field will be zero inside the circle, outside the circle I haven't got a clue.
    (b) For a circle with a uniform line charge it would have been simple, but I get the following result:
    [itex]\vec{E} = \frac{1}{4 \pi \epsilon_0}(\lambda \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\hat{r}}{r^2} dl' - \lambda \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \frac{\hat{r}}{r^2} dl') = \vec{0}[/itex]
    But this can't be correct, right? What am I doing wrong?
    (c) I really wouldn't know how to start answering this question. I guess it falls off according to [itex]\frac{1}{r^2}[/itex], but I don't know how to calculate this.

    I have my final exam tomorrow, any help would be greatly appreciated!
     
  2. jcsd
  3. Aug 4, 2011 #2

    kuruman

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    You need to draw yourself a picture. Include in your picture the vector contributions dE to the electric field from two diametrically opposed points, one with charge +dq and the other with charge -dq. Note that the two arrows do not point in the same direction. So you need to find an expression for the x-component dEx and integrate that to find Ex. Then do a separate calculation to find dEy and from it Ey.

    It would be a good idea to choose one of the axes, x or y, such that it goes through the boundaries where the distribution changes sign. This introduces some obvious symmetries about the principal planes, xy, yz and zx.
     
  4. Aug 4, 2011 #3
    Thanks for your answer!

    So, I tried to draw the contributions. Am I correct that inside the circle the dE due to the negative charge points toward the negative charge and dE due to the positive charge also points in that direction, thus doubling the field? Or is this too naive?

    For a point on the z axis, dE due to the positive charge points in the +z direction and towards the negative charge in the xy plane and dE due to the negative charge points in the -z direction also towards the negative charge in the xy plane. Is this correct?

    If I have the directions right I think I can calculate the integrals by myself.
    Thanks again for your help!
     
  5. Aug 4, 2011 #4

    kuruman

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    It is not too naive. If the x-axis is the line separating the positive and negative sides, the yz plane is a mirror plane. This means that the field cannot have a component perpendicular to the plane, i.e. in the x-direction.
    Also correct. The xy plane is also a mirror plane, therefore there can be no z-component.
    Good luck with your integrals.
     
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