To begin with, I'll have to assume you mean surplus electron charges on a sphere. Obviously, the integral of the inverse of distances between charges would be comparatively easy to relate to physical quantities since that's basically potential. So let's talk about this sphere, radius $R$, with a charge $Q$ on it, made up of $Q/e$ electrons. The self capacitance of a conducting sphere is $4 \pi \epsilon_0 R = R/k$. Then we can establish that the total potential energy is $\frac{1}{2}C V^2 = \frac{1}{2} \frac{Q^2}{C}$.
$$E=\frac{Q^2}{2 C} = k e^2 \frac{N^2 }{2 R} = k e^2 \sum_{i=1}^N \sum_{j=1}^{i-1} \frac{1}{|r_i-r_j|} $$
For the sum of distances (not the inverse), you can get a formula like this.
$$\sum_{i=1}^N \sum_{j=1}^{i-1} |r_i-r_j| = \frac{2}{3} N^2 R$$
This falls out from the calculus of the summation, just like the capacitance value. I didn't do that, instead I just wrote a code that discovered the relationship to my satisfaction.
program sphere
implicit none
double precision :: mu, theta
double precision :: mu2, theta2
double precision, dimension(3) :: r1, r2, rand
integer :: i, j, N
double precision :: thesum,thesum2, ind
double precision, parameter :: pi = 3.14159265
double precision :: d, rad
N = 5000
rad = 2.
ind = 0
thesum = 0.
thesum2 = 0.
do i = 1,N
r1 = random_points(rad)
do j = 1,i-1
ind = ind+1
r2 = random_points(rad)
d = sqrt(sum((r1-r2)**2))
thesum = thesum + 1./d
thesum2 = thesum2 + d
end do
end do
write(*,*) ' N= ',N,' number= ',ind
write(*,*) ' pot/N^2 ',thesum/N**2
write(*,*) ' len/N^2 ',thesum2/N**2
contains
function random_points(r)
implicit none
double precision, dimension(3) :: random_points
double precision, intent(in) :: r
double precision :: theta, mu
double precision, dimension(3) :: rand
call random_number(rand(1))
call random_number(rand(2))
theta = 2.*pi*rand(1)
mu = rand(2)*2.-1.
random_points(1) = cos(theta)*sqrt(1.-mu**2)*rad
random_points(2) = sin(theta)*sqrt(1.-mu**2)*rad
random_points(3) = mu*rad
end function random_points
end program sphere
I don't know if this value has any physical utility. To begin with, we can put it in terms of more familiar physical values.
$$\frac{2 N^2 R}{3} = \frac{2 Q^2 R}{3 e^2}$$
The problem with doing a distance integral is that I can't think of any physical thing for which this would matter. Field and potential are $1/r^2$ and $1/r$ and if you integrate again, you get $ln(r)$. I suppose some forces do grow with distance, and maybe they're proportional to distance.
I'm confident that the relationship between those two values require a relatively simple equation.
As long as the geometry is sufficiently simple, this will be true for many similar questions. That's in the domain of math.
EDIT:
I think that the revise problem is to constrain:
$$i=1 .. N$$
$$ |\vec{r}_i| < R$$
Then show that 1 being true implies 2 below
- the sum of $1/r$ integral of all the charges over the entire volume is constant
- the summed distance between all points is at a maximum
I think this could be what is being asked. It's a little lofty, but I'm sure it's entirely doable. My suspicion is that it would apply for any arbitrary region.
THE CALCULUS:
I'll present the way to get these numbers by integrating, partly because I think it would be helpful for an incoming freshman to see. The problem my above code solves is to integrate the values of $1/r$ and $r$ over all pairs of points over a sphere. When I do the actual integral I'll multiply it by 1/2 because the integral would otherwise double-count the pairs. Integrating is basically a way of using math to describe a problem with infinite points.
Let's start out. The surface area of the sphere is:
$$SA = 4 \pi R^2$$
The charge density is the number divided by the surface area. This is the number of charges per unit area on the surface of the sphere.
$$\sigma = \frac{N}{SA}$$
You can integrate over any variable you'd like, I'll chose the angle between the x-axis and the vector. I'm going to denote this with a prime to indicate that it is the "second" piont in the pair, and the "first" point in the pair will simply be fixed.
$$ \vec{r}' = <x',y',z'> = < R \cos{\theta}, 0, R \sin{\theta} >$$
$$ \vec{r} = < 1, 0, 0 >$$
Define the distance between them. This is a scalar.
$$ d = | \vec{r} - \vec{r}' | $$
I'm going to do a surface integral to cover all the $\vec{r}'$ and then do another surface integral (times 1/2) over all the $\vec{r}$. That's the jist of it, but there are lots of symmetries involved. These symmetries reduce the dimensionality of the $\vec{r}'$ integral by 1 and the $\vec{r}$ integral by 2. That means the latter isn't even an integral. The proposition behind these is that
- You can rotate the $\vec{r}'$ point around the x-axis and it doesn't change the distance
- You can move the $\vec{r}$ point all around the sphere and it doesn't change the distance
Now I'm at a point where I can write the integral. First for the total electrostatic energy. Note that $e \sigma$ is the charge density since I used $\sigma$ as the number density. The first expression of charge density time surface area is the multiplier that I use in lieu of an outer integral. The $2 \pi y'$ is needed to correctly use the x-axis symmetry, it equates to multiplying by the perimeter of a washer centered about the x-axis.
$$ N = (e \sigma SA) \frac{1}{2} \int_0^{\pi} 2 \pi y' \frac{k e \sigma}{d} d \theta = k e^2 \frac{ N^2}{2} $$
This is the result I wanted. The same manner is used almost identically to reproduce the number for the sum of distances between points. I'll leave out the charges because there's no clear physical interpretation.
$$ sum = ( \sigma SA) \frac{1}{2} \int_0^{\pi} 2 \pi y' \sigma d d \theta = \frac{2 N^2}{3} $$
And that's the integral. Using a computational algebra package is helpful, but everything should be sufficiently defined here.
