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Homework Help: Electromagnetism, potential

  1. Jan 26, 2006 #1

    Hi everyone :-) This is my first post here!
    I have a question regarding electrostatics, in the picture above. I am not looking for the solution at all as this work will be assesed, Im just hoping that someone can verify/ poke me in the right direction.

    First of all potential is q/(k*x) in x-hat direction?
    Integrate that over 2*Pi*a to get potential of a hoop.
    Using cosine rule to express x in terms of theta, then integrate over whole hoop from 0 to Pi..
    Thats what I have been trying, and Im not having any joy. Any small pointers would be most appreciated!

    I hope that was okay for a first post!

  2. jcsd
  3. Jan 26, 2006 #2
    You are almost correct. Wrong point is all potential is [tex]4\pi a^2 k \sigma / r = kq/r (or \frac q {4\pi \epsilon_0 r} )[/tex]

    First think of the ring portion on the sphere between [tex]\theta[/tex] and [tex]\theta + d \theta[/tex] (area is [tex]2\pi a^2 \sin \theta d \theta[/tex]), and calculate the electric field the ring creates at P (note you only need to calculate horizontal factor ignoring other directions.) Then integrate it over [tex]0 \leqq \theta \leqq \pi[/tex].

    That's all and I'm also calculating it.
    Last edited: Jan 26, 2006
  4. Jan 26, 2006 #3
    Hi, thanks for the reply :-)
    I am a little mystified by the area you mention... in what orientation are you taking this ring portion?
    Maybe it will come to me, I am working on the problem now but at the moment I cant see it....

  5. Jan 26, 2006 #4
    Actually, I see where you get the area from now :-) Ill keep at it..
  6. Jan 26, 2006 #5
    under construction :)

    ok. I've done all the calculations.

    First, the electric charge in the ring on the sphere between [tex]\theta and \theta + d \theta[/tex] is [tex]\2pi a^2 \sigma[/tex]. So the electric field at P this ring contributes is [tex]dE = \frac {2\pi a^2 \sigma \sin \theta d \theta} {x^2} \cos \angle QPO[/tex] (Q is the point on the sphere at [tex]\cos \theta[/tex]. You can take into account only the horizontal component of the electric field, so you need to multiply [tex]\cos \angle QPO[/tex]. So consider [tex]x^2 = a^2 + r^2 - 2ar \cos \theta[/tex] because of law of cosines,

    [tex]dE = \frac {2\pi a^2 \sigma \sin \theta } {x^2} \cdot \frac {r-a \cos \theta} { x} = 2 \pi a^2 \sigma (r^2+a^2 - 2ar \cos \theta)^{-3/2} (r-a \cos \theta) \sin \theta d \theta[/tex]

    We need to integrate [tex]0 \leqq \theta \leqq \pi[/tex], but substitute [tex]u = \cos \theta, [/tex] so [tex] d \theta = \frac {d \theta}{du} du = - \frac 1 {\sin \theta} du[/tex].

    So we integrate:

    [tex]dE = 2 \pi a^2 \sigma (r^2+a^2 -2aru)^{-3/2}(r-au)du[/tex] over [tex]-1 \leqq u \leqq 1[/tex] (note sign is reversed.)

    Using partial integral,

    [tex]E = 2\pi a^2 \sigma (\frac 1 {ar} (r^2+a^2 -2aru)^{-1/2} (r-au) \vert_{-1}^1 + \frac 1 {ar} \int_{-1}^1 a(r^2+a^2-2aru)^{-1/2} du)[/tex]

    [tex]= 2\pi a^2 \sigma (\frac 1{ar} (\frac {|r-a|}{|r-a|} - \frac {|r+a|} {|r+a| } ) + \frac 1{ar} (-\frac {2a}{2ar} (r^2+a^2-2aru)^{1/2} \vert_{-1}^1)[/tex]

    [tex]= 2\pi a^2 \sigma(\frac 1{ar} \frac 1 r(-(|r-a|-|r+a|))[/tex]
    [tex]= \frac {4 \pi a^2 \sigma} {r^2}
    Last edited: Jan 26, 2006
  7. Jan 26, 2006 #6
    Hiya, thanks for the effort you have put in!
    I think I have the solution now, I found the potential first and then the electric field from that, and everything drops out nicely and appears like a point charge as required. I will of course work through your solution also to compare, I thought mine was correct although its a lot shorter than yours...Ill have to check I havent missed anything!


  8. Jan 26, 2006 #7
    oh yeah, it says calculate potential first! lol.
  9. Jan 26, 2006 #8
    Yeah! If you calculate potential, you don't have to consider [tex]\cos \angle QPO[/tex], so it's much shorter and easier! This cosine part makes the calculation complicated, because of this part, partial integral is necessary.
  10. Jan 27, 2006 #9
    My calculation for potential is (as the question requests :))

    [tex]d \phi = \frac {2\pi a^2 \sigma \sin \theta} x d \theta[/tex]

    so it follows that:

    [tex]\phi = \int_0^\pi \frac {2 \pi a^2 \sigma \sin \theta} x d \theta [/tex]

    [tex]= \int_0^\pi 2\pi a^2 \sigma (r^2+a^2 -2ar \cos \theta)^{-1/2} \sin \theta d \theta [/tex]

    [tex]= \int_{-1}^1 2 \pi a^2 \sigma (r^2+a^2 - 2aru)^{-1/2} du = \frac {4 \pi \sigma a^2} r[/tex]

    actually much easier...
    Last edited: Jan 27, 2006
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