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Field of non-uniformly charged ring

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A thin rubber ring of radius R lies in the xy-plane and is centered at the origin. The ring carries line charge density λ=λo*sin(Ø), with a constant λo with units C/m. define tan(Ø)=y/x. Calculate the magnitude and direction of the electric field E(z) created by the ring on points on the positive z axis.



    2. Relevant equations

    E=kq/r^2

    3. The attempt at a solution

    Symmetry implies only the component Ez will be nonzero along the z axis.
    Desiring to integrate over Ø, my limits of integration will be 0 to pi.

    Now dE=k*dq/r^2 with r=(R^2+z^2).

    Q/R=λo*sin(Ø) so dQ=R*λo*cos(Ø) dØ

    and dEz= k*R*λo*cos(Ø) dØ/(R^2+z^2)

    however, integrating this over Ø results in multiplying everything by sin(pi)-sin(0) = 0, so clearly I've made at least one mistake here.

    All help is appreciated!
     
    Last edited: Sep 21, 2009
  2. jcsd
  3. Sep 21, 2009 #2
    the result must be zero for the sake of symmetri. Your result is correct (I think)
     
  4. Sep 21, 2009 #3
    Symmetry in the x and y directions makes sense, as an equal amount of ring and angle phi are distributed to both the positive and negative halves. However the field would certainly have a Z component at points along the Z axis, as the full ring exerts charge in either a positive or negative z direction.
     
  5. Sep 21, 2009 #4
    the charge is ditributed in iqual amount in each semi-circle but with defferent signs, so symmetry implies that the z component must be zero. However I now realize that there must be a component in the x-y plane
     
  6. Sep 21, 2009 #5

    gabbagabbahey

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    I think you should draw a picture...the angle [tex]\phi[/itex] is measured entirely in the xy-plane, and any point on the ring is given by [itex]x=R\cos\phi[/itex], [itex]y=R\sin\phi[/itex] and [itex]z=0[/itex].

    If the ring is an entire circle, the limits on [itex]\phi[/itex] will be zero to [itex]2\pi[/itex]

    Not quite. By definition of linear charge density, [itex]dq=\lambda ds=\lambda_0 \sin\phi Rd\phi[/itex]

    No, not only should the cosine be a sine, but you haven't properly selected the z-component of [itex]d\textbf{E}[/itex].

    Again, draw a picture...the field [itex]d\textbf{E}[/itex] from a tiny piece of the ring points from that piece, to the point on the z-axis at which it is measured...you want to find the z-component of that vector.

    Even after you do the calculation properly, you will still find that [itex]E_z=0[/itex], however, if you also calculate the other components you should find that [itex]E_y\neq0[/itex].
     
  7. Sep 22, 2009 #6
    Thanks. After figuring out I had the wrong component, I was able to solve the problem.
     
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