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Electric field of a non-uniformly charged ring

  1. Feb 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the magnitude and direction of the electric field due to the ring in the point P on the z-axis.
    06b46043ae.jpg

    2. Relevant equations
    Charge distribution on the ring is given by [itex]λ(φ)=λ_0\cdot sin(φ)[/itex]. This should result in the total charge on the ring being zero.
    Electric field is given by [itex]E=\frac{1}{4\pi\epsilon_{0}}\int_{Path}\frac{\lambda\left(r\right)}{r^{2}}\hat{r}dl[/itex]
    The distance [itex]r[/itex] from any point on the ring to the point P: [itex]r=\sqrt{z_0^2+R^2}[/itex]

    3. The attempt at a solution
    Due to symmetry and the non-uniformity of the charge distribution we can say that the electric field in the z-direction is 0 ([itex]E_z=0[/itex]) but there will be an x-component as seen in the drawing I made.
    64baaffdd6.jpg
    So now to select the x-component we say [itex]dE=\frac{dE_{x}}{\sin\left(\theta\right)}[/itex] so [itex]dE_x=dE\cdot\sin\left(\theta\right)[/itex].
    [tex]
    \begin{eqnarray*}

    dE_{x} & = & dE\cdot\sin\left(\theta\right)\\

    & = & \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)dq\\

    & = & \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda dl\\

    & = & \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda_{0}\sin\left(\varphi\right)dl\\

    & = & \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda_{0}\sin\left(\varphi\right)Rd\varphi\\

    & = & \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\frac{R}{r}\lambda_{0}\sin\left(\varphi\right)Rd\varphi\\

    & = & \frac{\lambda_{0}R^{2}}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{3}}\sin\left(\varphi\right)d\varphi\\

    & = & \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^2}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\sin\left(\varphi\right)d\varphi

    \end{eqnarray*}
    [/tex]

    Then you should integrate to get the final answer [tex]
    \begin{eqnarray*}

    E_{x} & = & \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\int_{0}^{2\pi}\sin\left(\varphi\right)d\varphi\\

    & = & 0

    \end{eqnarray*}
    [/tex]

    But as you can see that results in 0 so I'm not really sure where the mistake is but I think it's either in selecting the x-component or I'm doing the integration wrong.
     
  2. jcsd
  3. Feb 6, 2016 #2

    haruspex

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    What makes you think the field will be nonzero in the x direction? Have you looked at the symmetry for that? Any other directions to consider?
     
  4. Feb 6, 2016 #3
    Because of the charge distribution there will be a negatively charged lower ring ([itex]φ=π[/itex] to [itex]φ=2π[/itex]) and a positively charged upper ring ([itex]φ=0[/itex] to [itex]φ=π[/itex]). The negative semi-ring will have a negative z-component; a positive x-component and a negative y-component (?) and the positive semi-ring will have a positive z-component; a positive x-component and a negative y-component on the z-axis thus the z-components cancel out and we're left with the x and y components?
    Or because of the fact [itex]λ=λ_0\cdot \sin(φ)[/itex] which is 0 at both [itex]φ=0[/itex] and [itex]φ=π[/itex] there can be no component in the x-direction and we're only left with the y-component?
     
  5. Feb 6, 2016 #4

    haruspex

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    Let S' be the reflection of S in the XZ plane, so in terms of the diagram its angle is -φ. What is the direction of the net field at P due to the elements at S and S'?
     
  6. Feb 6, 2016 #5
    Okay so an element at [itex]S[/itex] produces an electric field in point P like [itex]E=-E_x \hat{x}-E_y \hat{y}+E_z \hat{z} [/itex] and an element at [itex]S'[/itex] produces [itex]E=+E_x \hat{x}-E_y \hat{y}-E_z \hat{z} [/itex] thus adding both of those we have a total electric field of [itex]E=-2E_y \hat{y} [/itex]?
     
  7. Feb 6, 2016 #6

    haruspex

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    Yes.
     
  8. Feb 7, 2016 #7
    Now when I try to calculate the electric field in the y-direction I get [itex]E_{y} = \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\int_{0}^{\pi}\sin\left(\varphi\right)d\varphi[/itex] (following the same procedure as I did in the first post) my question would be what to pick for the integration limits. If I chose [itex]φ=[0,π][/itex] then the integral gives me [itex]2[/itex] but if I pick [itex]φ=[π,2π][/itex] then I get [itex]-2[/itex]. But I'm guessing since the field has to be in the negative y-direction at the end we have to pick [itex]φ=[0,π][/itex] as our limits such that when we multiply by the -2 we got from symmetry arguments the final answer is negative?
     
  9. Feb 7, 2016 #8

    haruspex

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    Your integral is wrong.
    I had not previously studied the integral you did for the x component in the initial post. Your expression for Ex there is not correct. Esin(θ) gives you the component of E in tne XY plane, but for the component in specifically the X or Y direction there must be some dependence on φ.
     
  10. Feb 7, 2016 #9
    So it should be [itex]dE_y=\sin(φ)\cos(θ)dE \hat{y}[/itex] then? Because that would be the y-component of a vector in spherical coordinates.
     
  11. Feb 7, 2016 #10

    haruspex

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    Not sure I understand your notation there, but yes, the component factor you need is cos(θ)sin(φ).
     
  12. Feb 7, 2016 #11
    Oh what I tried to say was that an element of the electric field in the y direction was equal to the y-component of the electric field vector.

    Sorry if I'm a bit slow to understand but the resulting electric field would be [itex]dE_y=\frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\cos\left(\theta\right)\cdot \lambda_{0}\sin^2\left(\varphi\right)\cdot Rd\varphi[/itex] which then leads to [itex]E_y=\frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{z_{0}R}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\int_{0}^{2\pi}\sin^{2}\left(\varphi\right)d\varphi[/itex] and [itex]E=-2\cdot E_y[/itex] as per symmetry arguments.
    But then another question arises from this, if [itex]z_0\gg R[/itex] this configuration should reduce to a dipole which has an electric field that is proportional to [itex]z_0^{-3}[/itex] in this case but the electric field I found is proportional with [itex]z_0^{-2}[/itex] for [itex]z_0\gg R[/itex].
     
  13. Feb 7, 2016 #12

    haruspex

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    Sorry, I copied your error without noticing. Originally you wrote sin(θ), which was correct, not cos.
     
  14. Feb 7, 2016 #13
    Okay so then it becomes [itex]dE_y=\frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda_{0}\sin^{2}\left(\varphi\right)Rd\varphi[/itex] which then results in [tex]
    \begin{eqnarray*}

    E_{y} & = & \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\int_{0}^{2\pi}\sin^{2}\left(\varphi\right)d\varphi\\

    & = & \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\cdot\pi\\

    E & = & -2\cdot E_{y}\hat{y}\\

    & = & -\frac{\lambda_{0}}{2\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\hat{y}

    \end{eqnarray*}


    [/tex]

    This does properly reduce to a [itex]z_0^{-3}[/itex] proportionality for large [itex]z_0[/itex] so I'm guessing this is correct?
     
  15. Feb 7, 2016 #14

    haruspex

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    Yes, I think it's right.
     
  16. Feb 7, 2016 #15
    Thank you so much for all your help!
     
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