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D dimension scalar potential for point charge

  • #1

Homework Statement


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]


Homework Equations





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]

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
383
0

The Attempt at a Solution



The electric field strength is known to be:

...

[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]

[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]

...

Why am I getting an extra [itex]\frac{1}{\sqrt{d}}?[/itex]
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:
 
  • #3
How did you go the first equation above to the second?
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]

what is it that you are summing over?
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:
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
6
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.
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]
 
  • #5
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]
That was indeed what I was trying to argue. I feel quite silly now. Thankyou for your help
 

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