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Field due to ring charge at x-y axis.

  1. May 3, 2017 #1
    1. The problem statement, all variables and given/known data

    A charge ##Q## is distributed unifromly around a thin ring of radius ##b## which lies in ##xy## plane with its centre at the origin. Locate the point on the positive ##z## axis where the electric field is strongest ?

    Physics ring charge.png

    2. Relevant equations



    3. The attempt at a solution

    ##\displaystyle d\vec E = {dQ \over b^2 + r^2 }\cos \alpha##. ##r## is the distance between the point, ##\alpha## is the angle between ##z## axis and the displacement vector , and the ring along ##z## axis.

    ##\displaystyle d\vec E = {\lambda dl \over b^2 + r^2 } \cos \alpha = d\vec E = {\lambda bd\theta \over b^2 + r^2 } \cos \alpha##.

    Using ##\displaystyle \cos \alpha = {r \over \sqrt{b^2 + r^2}}##.

    ##\displaystyle E = {br\lambda \over (b^2 + r^2)^{3/2} } \int^{2\pi}_{0} d\theta = {2\pi br\lambda \over (b^2 + r^2)^{3/2} }##

    Since ##\displaystyle \lambda = {Q\over 2\pi b}##,

    I get,

    ##\displaystyle E = {Qr \over (b^2 + r^2)^{3/2}}##

    Which has a maximum ##\displaystyle r = {2\over 3\sqrt{3}b^2}##.

    Am I correct ?
     
    Last edited: May 3, 2017
  2. jcsd
  3. May 3, 2017 #2

    kuruman

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    The derivation of the electric field is correct. Your final answer is not. If b is the radius, your value for r has the wrong dimensions. It must proportional to b not to b-2.
     
  4. May 3, 2017 #3
    Oh that was WA's mistake. I got lazy and let it calculate maxima. Is ## r = \pm b/ \sqrt{2}## correct ?
     
  5. May 3, 2017 #4

    kuruman

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    Where is b on the right side of the equation?
     
  6. May 3, 2017 #5
    sorry.

    I edited my post. Is the answer ##\pm b/\sqrt{2}## ?
     
  7. May 3, 2017 #6

    kuruman

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    I would pick the positive value because the problem specifies "on the positive z axis". Otherwise it's fine.
     
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