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Charge density in terms of (r,θ) but need it in terms of the vector r'

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A sphere of radius R, centered at the origin, carries charge density
    ρ(r,θ) = (kR/r2)(R - 2r)sinθ,
    where k is a constant, and r, θ are the usual spherical coordinates.
    Find the approximate potential for points on the z axis, far from the sphere.


    2. Relevant equations
    potential of the dipole
    V(r,θ) = 1/4πεo * 1/r2 ∫ r'cosθ' ρ(r') dτ'


    3. The attempt at a solution
    I just want to know why I'm giving the charge density in terms of r and θ and yet I need it in terms of r'. I know that the vector r' is related to θ' by r' = rcosθ'
    So I tried to say that since θ is the inclination from the xy plane and θ' is the angle between r and r' so that θ'+θ = π/2. Then that would give me
    sinθ = [(π/2))-θ'] = cosθ'
    Since the direction of r' is cosθ' I thought maybe I could replace r with r'?
    so maybe
    ρ(r',θ') = (kR/r'2)(R - 2r')cosθ' so
    ρ(r') = (kR/r'2)(R - 2r') ?????
    PLEASE help I have a test on this tomorrow..
     
    Last edited: Nov 8, 2009
  2. jcsd
  3. Nov 8, 2009 #2
    Looks to me that you are confusing yourself on a simple matter. When inside the integral, it is often convenient to use primed-coordinates to separate from the original matter. If you had

    [tex]
    V(r,\theta)=\frac{1}{4\pi\varepsilon_0}\,\frac{1}{r^2}\int r\cos[\theta]\rho(r)d\tau
    [/tex]

    it might confuse you because your limits of integration would also be in terms of [itex]r[/itex] and [itex] \theta[/itex]. The prime markers just show that, inside the integrand, [itex]r'[/itex] is slightly different from [itex]r[/itex], but does become [itex]r[/itex] after integration; this way you have

    [tex]
    V(r,\theta)=\frac{1}{4\pi\varepsilon_0}\,\frac{1}{r^2}\int r'\cos[\theta']\rho(r')d\tau'
    [/tex]

    to tell the difference between what needs to be integrated and what is already taken care of through integration.
     
  4. Nov 8, 2009 #3
    so in ρ(r,θ) = (kR/r2)(R - 2r)sinθ, r and θ are bascially r' and θ'?
    when I integrate it that way I'm getting
    ∫ r'cosθ' kR(R-2r')sin2θ' dr'dθ'dφ'
    then integrating ∫sin2θ' cosθ' dθ' I get
    1/3sin3θ' (0 -> pi) which gives me zero.
    And the answer is supposed to be
    V(z) = 1/4πεo kπ2R5/(24z2)
     
  5. Nov 8, 2009 #4

    gabbagabbahey

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    What makes you think that?...I also get that the monopole and dipole terms both vanish, and so the dominant term will be the quadrapole term....
     
  6. Nov 8, 2009 #5
    But then the answer I'm supposed to be aiming for is wrong? At least I know maybe my qudrapole term wasn't wrong then, and maybe it's right?
    3πkR5/64εoz3 ?
    I got the anwser that it is suppose to be of this website http://einstein1.byu.edu/~masong/emsite/S2Q50/S2Q50.html
     
    Last edited by a moderator: Apr 24, 2017
  7. Nov 8, 2009 #6

    gabbagabbahey

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    That's a pretty sketchy website to be looking for homework solutions...the answer they give doesn't even have the correct units, so that should be a dead giveaway that something is wrong.

    Your quadrapole term is a little off....why not show me your calculation?
     
  8. Nov 8, 2009 #7
    Here goes:

    V(r,θ) = 1/4πεo * 1/r3 ∫∫∫0R (3/2 cos2θ' - 1/2) kRsin2θ' (Rr'2 - 2r'3) dr'dθ'dφ'

    = 1/4πεo * 1/r3 ∫∫0π (3/2 cos2θ' - 1/2) kRsin2θ' (-1/6 kR4) dθ'dφ'

    | Using 3/2 cos2θ'(cos2θ' - 1)
    | = 3/2 (cos4θ' - cos2θ')
    | = 3/2 [(1/2 + 1/2cos2θ')2 - cos2θ']

    = 1/4πεo * 1/r3 ∫∫0π [3/2 [(1/2 + 1/2cos2θ')2 - cos2θ'] sin2θ' (-1/6 kR4) dθ'dφ'

    = 1/4πεo * 1/r3 ∫∫0π 3/2[(1/4 + 1/2cos2θ' + 1/4cos22θ')- cos2θ'] -1/2sin2θ' (-1/6 kR4) dθ'dφ'

    = 1/4πεo * 1/r3 (-1/6 kR4) ∫0( 3/2[(θ'/4 + 1/4sin2θ' + θ'/8 + 1/16sin2θ')] - 1/2(θ'/2 - 1/4sin2θ'|0π dφ'

    | Since all the sine terms are zero I got

    = 1/4πεo * 1/r3 (-1/6 kR4) ∫0( 3/2[(θ'/4 + θ'/8)] - 1/2(θ'/2)|0π dφ'

    = 1/4πεo * 1/r3 (-1/6 kR4) ∫0 3/2(-3/8π) dφ'

    = 1/4πεo * 1/r3 (-1/6 kR4) 3/2(-3/8π) * 2π

    = 3πkR5 / 64εor3 |r=z

    = 3πkR5 / 64εoz3



    I think there may be a mistake in my use of the half-angle identity?
     
  9. Nov 8, 2009 #8

    gabbagabbahey

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    Why are you using this?

    [tex]\left(\frac{3}{2}\cos^2\theta'-\frac{1}{2}\right)=\frac{1}{2}(3\cos^2\theta'-1)=\frac{1}{2}[3(1-\sin^2\theta')-1]=1-\frac{3}{2}\sin^2\theta'\neq\frac{3}{2}\cos^2\theta'(\cos^2\theta'-1)[/tex]
     
  10. Nov 8, 2009 #9
    Because I was distributing the sin2 to just that term and then trying to solve that integral. But I see that you're way is much easier. Thanks!
     
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