# D dimension scalar potential for point charge

1. Dec 1, 2009

### babyEigenshep

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

2. Relevant equations

3. The attempt at a solution

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

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

2. Relevant equations

3. The attempt at a solution

2. Dec 1, 2009

### jdwood983

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 $i$ index on any of your terms

PS: to get scalable parenthesis, use the command \left( and \right) between your work

3. Dec 1, 2009

### babyEigenshep

The first equation is
$$E(r) = \sqrt{\frac{1}{d}\sum_{i=1}^d\left(\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}\right)^2}$$

taking $\frac{1}{d}$ through the sum yields

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

which is the same as

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

so we can collect it into 1 squared term

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

I am summing over the dimensions. By definition of $E(r)$ as the magnitude of the electric field,
$$E(r) = \sqrt{\sum_{i=1}^d \left. E_{x^i}\right. ^2$$
where $E_{x^i}$ are the components of $\vec{E}$
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
4. Dec 1, 2009

### gabbagabbahey

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 $1+2+3=2+2+2$

5. Dec 1, 2009

### babyEigenshep

That was indeed what I was trying to argue. I feel quite silly now. Thankyou for your help