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Electric Field of a Charged Ring

  1. Oct 15, 2011 #1
    Hi everyone,

    1. The problem statement, all variables and given/known data
    Ring.jpg
    Let a charge Q be uniformly distributed on a circular ring defined by a < [itex]\rho[/itex] < b. Find D at (0,0,h).


    2. Relevant equations
    E = kQ/r2 ar
    D = [itex]\epsilon[/itex]o E

    3. The attempt at a solution

    Well I thought I had this figured out, but I was wrong and I still can't figure it out after 2 hours :(

    Since I have a range for the radius (a and b), I thought that I could calculate the E-field for both a and b separately then simply apply b-a to the field.

    Finding both r1 and r2:
    cos ∅1 = h / r1 = h / [itex]\sqrt{a^2 + h^2}[/itex]
    cos ∅2 = h / r2 = h / [itex]\sqrt{b^2 + h^2}[/itex]

    Then, E = kQh / r^2 [ cos ∅1 - cos ∅2 ]

    I just don't know how to tackle this one. I have a large gap that I need to make note of.

    The solution in the back of the book is D = [itex]\frac{Qh}{2\pi(b^2-a^2)}[/itex] [[itex]\frac{1}{\sqrt{a^2+h^2}}[/itex] - [itex]\frac{1}{\sqrt{b^2+h^2}}[/itex] ]

    I would greatly appreciate any help!! I kind of think I had the right idea, I just don't understand how to fix what I've done wrong.

    Any help is greatly appreciated!! :)
     
  2. jcsd
  3. Oct 16, 2011 #2
    First you need to calculate the charge using the charge density. We know that:
    [itex]Q = \int_{a}^{b}\int_{0}^{2\pi}\rho_lrd\phi[/itex]

    From there you should find the potential of a disk along the z-axis. The integral should be set up as:
    [itex]d\phi = \frac{\rho_l*2*\pi*rdr}{4*pi*\epsilon_o\sqrt{z^2+r^2}}[/itex].

    If you integrate correctly it should come out to be in terms of both b and a. Then using the fact that [itex]E =-\nabla\phi[/itex] in the z-direction should get you to the answer. Let me know if you need more explicit help.
     
  4. Oct 17, 2011 #3
    Thanks for your response :)

    A few questions though,

    Charge enclosed: Qenc= ρsdS
    I don't understand how I can solve for this when I don't know what ρs or Qenc are.

    Also I don't understand your last comment: E=−∇ϕ
    Did you mean to use E=−∇V (where V is the potential)?

    As always, thanks for the help :)
     
  5. Oct 17, 2011 #4
    That is true you don't know what [itex]\rho_s[/itex] or [itex]Q_{enc}[/itex] is but you do know that there is going to be a charge distribution on the ring. You know that it will have a total charge Q enclosed. If you recall from your solution, it is indeed in terms of Q which is the charge on the ring.

    So you want to integrate [itex]Q = \int_{a}^{b}\int_{0}^{2\pi}\rho_srd\phi[/itex] and solve for [itex]\rho_s[/itex] in terms of Q.

    And sorry about the notation for [itex]E =-\nabla\phi[/itex] in this case I did mean that [itex]\phi[/itex] was the potential and that's just the notation I'm used to. So yes, [itex]\phi[/itex] is the same as V.

    Also, when you do [itex]E =-\nabla\phi[/itex], you want to solve for when z = h.

    If you have any more questions let me know.
     
  6. Oct 17, 2011 #5
    Qenc= ρsdS
    Qencs2∏*[itex]\frac{1}{2}[/itex](b2-a2)
    ρs=Qenc/∏(b2-a2)

    So now I'd have to find resultant vector R (I'm bad at this). Here goes though:
    If I drew a triangle with a radius of r and a height of h I would get: R=(r2+h2)1/2. But now I'd have to convert this to cylindrical coordinates, so I would replace h with z and r with ??? (I don't exactly know how to do this here because I don't have a value for the radius - I have 2 possible values. I'm over-complicating it in my head. I don't really understand what radius I need/want.)

    Then after that, I can simply plug it into the formula for potential.

    Because it's a point charge, the potential is simply:

    V=[itex]\frac{Q}{4\piεR}[/itex]

    After I find the potential, I can apply [itex]E =-∇V[/itex], then apply [itex]D =εE[/itex]

    Also, why is it that we sometimes use h (the length) and sometimes we refer to it as z? This really confuses me when I try to write my vectors going from cartesian to cylindrical.

    Thank you so much for your help runnergirl :D
     
  7. Oct 17, 2011 #6
    So you're close. In this case we're integrating over r so instead of [itex]R = (r^2+h^2)^{1/2}[/itex] you'll use [itex]R = (r^2+z^2)^{1/2}[/itex]. This is because you're not solving for z = h yet. So when it's written like this, it is in cylindrical coordinates (a function of r and z). So yes, use a triangle with r being the change in the radial direction, which I'll show in a minute is what you integrate over, and z since that too is a variable in which you will use later.

    So now if you plug it all in:
    [itex]V = \int_{a}^{b}\frac{Q}{2\pi\epsilon_o(b^2-a^2)}\frac{rdr}{\sqrt{z^2+r^2}}[/itex]

    It's a fairly nice integral with a integral table. From there you do indeed apply [itex]E = -\nabla V[/itex] where in this case it will be only in the z-direction and solving for when z = h. So more explicitly:
    [itex]E = -\frac{\partial V}{\partial z}\mid_{z=h}[/itex].

    Well typically it is usually z that is used because most of the time in problems like these we're looking for the E-field along the z-axis which z is a variable. In this case, the reason it is h at the end is because you're evaluating the E(0,0,h) on the z-axis, so you're looking for the E-field at a particular point along the z-axis.

    Hope that helps.
     
  8. Oct 18, 2011 #7
    So out of curiosity, since this was a point charge:

    V=[itex]\frac{Q}{4\piεR}[/itex]

    But is it also fair to say that we could have used V = - ∫E dl ? I'm just curious as to why we're integrating to find potential because technically it is a point charge and we don't have to.

    This problem was a disaster and you've cleared it up for me. Thank you so much.
     
  9. Oct 18, 2011 #8
    Well since this is a charge Q distributed over the entire ring, it cannot be thought of as a point charge, unless you're looking at the far field. If z>>a,b the potential will due to the disk charge Q will behave like a point charge. So in this case, it is easier to find the potential, V (and we have to integrate over the entire ring), and then use [itex]E = -\nabla V[/itex] in the z-direction because the field is expected to be along the z-axis (this is due to the cylindrical symmetry of the problem). Now if we were given the E-field right away and asked to find the potential, then yes applying [itex]V = -\int E\cdot dl[/itex] would be a good method.

    Glad it makes more sense to you.
     
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