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D dimension scalar potential for point charge

  1. Dec 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that with [itex]d[/itex] spatial dimensions the potential [itex]\phi[/itex] due to a point charge [itex]q[/itex] is given by
    [tex]\phi (r) = \frac{\Gamma(\frac{d}{2}-1)}{4\pi^{d/2}}\frac{q}{r^{d-2}}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    The electric field strength is known to be:
    [tex]E(r) = \frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}[/tex]
    [tex]E(r) = \sqrt{(\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}})^2}[/tex]
    [tex]E(r) = \sqrt{\frac{1}{d}\sum_{i=1}^d(\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}})^2}[/tex]
    [tex]E(r) = \sqrt{\sum_{i=1}^d(\frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}})^2}[/tex]
    Since the magnetic field will be the same in all directions, we know that
    [tex]\vec {E}(r) = (\frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}},\frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}},\cdots)[/tex]
    [tex]\phi(r) = -\int\frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}dr[/tex]
    [tex]\phi(r) = \frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{1}{(d-2)}\frac{q}{r^{d-2}}[/tex]
    [tex]\phi(r) = \frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{1}{2(\frac{1}{2}d-1)}\frac{q}{r^{d-2}}[/tex]
    [tex]\phi(r) = \frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2}-1)}{4 \pi^{d/2}}\frac{q}{r^{d-2}}[/tex]

    Why am I getting an extra [itex]\frac{1}{\sqrt{d}}?[/itex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 1, 2009 #2
    How did you go the first equation above to the second? And also, what is it that you are summing over? You have the summation sign, but no [itex]i[/itex] index on any of your terms


    PS: to get scalable parenthesis, use the command \left( and \right) between your work :wink:
     
  4. Dec 1, 2009 #3
    The first equation is
    [tex]
    E(r) = \sqrt{\frac{1}{d}\sum_{i=1}^d\left(\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}\right)^2}
    [/tex]

    taking [itex]\frac{1}{d}[/itex] through the sum yields

    [tex]
    E(r) = \sqrt{\sum_{i=1}^d\frac{1}{d} \left(\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}\right)^2}
    [/tex]

    which is the same as

    [tex]
    E(r) = \sqrt{\sum_{i=1}^d \left(\frac{1}{\sqrt{d}}\right)^2 \left(\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}\right)^2}
    [/tex]

    so we can collect it into 1 squared term

    [tex]
    E(r) = \sqrt{\sum_{i=1}^d\left(\frac{1}{\sqrt{d}} \frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}\right)^2}
    [/tex]

    I am summing over the dimensions. By definition of [itex] E(r) [/itex] as the magnitude of the electric field,
    [tex] E(r) = \sqrt{\sum_{i=1}^d \left. E_{x^i}\right. ^2 [/tex]
    where [itex] E_{x^i}[/itex] are the components of [itex] \vec{E} [/itex]
    The sum is used with the fact that changing the order of the dimensions should not change the electric field to calculate the electric field from the magnitude.
     
    Last edited: Dec 1, 2009
  5. Dec 1, 2009 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    It almost looks like you are using this to argue that [tex]E_{x^i}=\frac{1}{\sqrt{d}} \frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}[/itex]....but you can't really conclude this from the fact that [tex] \sqrt{\sum_{i=1}^d\left(\frac{1}{\sqrt{d}} \frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}\right)^2}= \sqrt{\sum_{i=1}^d \left. E_{x^i}\right. ^2}[/itex] any more than you can conclude that 2=1, 2=2 and 2=3 from the fact that [itex]1+2+3=2+2+2[/itex]
     
  6. Dec 1, 2009 #5
    That was indeed what I was trying to argue. I feel quite silly now. Thankyou for your help
     
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