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Electric field with dielectric

  1. Feb 26, 2014 #1
    1. The problem statement, all variables and given/known data
    There's a ring of radius R with linear uniform charge density [tex]\lambda=constant[/tex] (which includes both free charges and polarization charges). The ring is in a dielectric region with [tex]\varepsilon_r[/tex]. I have to find the electric field on any point of the axis of the ring
    2. Relevant equations
    [tex]\vec E=\int \frac{\lambda dl (\vec r - \vec r')}{| \vec r - \vec r' | ^3}[/tex]


    3. The attempt at a solution

    I actually don't know and coulnd't find anywhere if I can calculate the electric field using the above formula when [tex]\lambda[/tex] includes free AND polarization charges.
     
  2. jcsd
  3. Feb 26, 2014 #2

    BvU

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    Is there a coefficient missing in your equation ?
     
  4. Feb 26, 2014 #3
    No, that's all I have. Since I'm given a linear density for the ring which includes free and polarization charges, I assume it's a dielectric, but I don't know its coefficient. With that density which includes polarization charges, is it ok to calculate the electric field, like in the formula I wrote before?
     
  5. Feb 26, 2014 #4
    Since the ring is submerged in a dielectric region with εr, the Electric field decreases by a factor of εr.
    So E' = E/εr, where E is the magnitude of the field without the dielectric, and E' with dielectric.
    Now take an infinitely small section on the circumference on the ring, dl. The charge associated is dq=λdl.
    The field associated with dq is 1/4piε (dq/r^2). However, since the ring is symmetric, the x-component of the net electric field, which is perpendicular to the axis, cancels out. Now all you need to consider is the y component. The y component of the field is Ey= E'cos(theta), where theta is the angle made between the axis, and the field vector coming from dq. Set "z" as the distance along the axis from the center of the circle. You can find cos(theta) using a relationship between "z" and "R", with the use of pythagorean theorem. Also, when you intergrate with respect to dl, change dl to Rd(theta). Integrate all contributions of dq from 0 to 2pi (because it is a circle) and you should be able to find the electric field anywhere along the axis. I understand that this is a sloppy description, but I hope this help anyways.
     
  6. Feb 26, 2014 #5
    It should be fine to calculate it treating it as free charges, because your density does include polarization charges.
     
  7. Feb 27, 2014 #6

    BvU

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    Strange. i seem to remember that
    [tex]\vec E= {1\over 4\pi\epsilon_0}\int \frac{\lambda dl (\vec r - \vec r')}{| \vec r - \vec r' | ^3}[/tex]in vacuum. In a medium it would be ##{1\over 4\pi\epsilon}## with ##\epsilon=\epsilon_0 \epsilon_r##, but I could be mistaken. So check.
    There is nothing in ##\lambda## that has to do with stuff outside the ring. No change in ##\lambda##, whatever ##\epsilon_r##.
     
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