Surface Integral

I want to estimate the energy of point charges over a sphere, however I think I am doing the integral wrong.

I know the energy is calculated by:

\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N} \frac{q_{i}q_{j}}{|p_{i}-p_{j}|}

Where pi are the position vectors and qi are the charges of the points. Since the charge I am taking is one, and the radius of my sphere is one, when I model the point charges as a uniform charge over the sphere I can get the energy to be:

\frac{1}{2}N^2\frac{1}{16 \pi^2} \int_{S}\int_{S} \frac{1}{|p-p_{0}|}dSdS

and then taking polar coordinates I get and fixing

p_{0}

(0,0,1)

I get:

\frac{N^{2}}{32 \pi^2}\int_0^{2\pi}\int_{0}^{\pi}\frac{sin(\theta)}{\sqrt{2-2cos(\theta)}}d\theta d\phi

which cancels to:

\frac{1}{2}N^2\frac{1}{4\pi}

However I know it should be

\frac{1}{2}N^2

, what am I doing wrong??

Reply 1

ghostwalker

Original post by adie_raz

However I know it should be

\frac{1}{2}N^2

, what am I doing wrong??

So you're out by a factor of 4pi, which happens to be the surface area of a sphere of radius 1. So, do you need to take charge density into account? Not that I know anything about this.

(edited 11 years ago)

Reply 2

adie_raz

Original post by ghostwalker

So you're out by a factor of 4pi, which happens to be the surface area of a sphere of radius 1. So, do you need to take charge density into account? Not that I know anything about this.

I did realise that I was out by the surface area which made me think I was doing it slightly wrong but am definitely close. I shall look into the charge density so thanks you for that.

I am unsure of my notation as well however, I am not sure if it should in fact be:

\frac{1}{2}N^2\frac{1}{16 \pi^2} \int_{S} \int_{S} \frac{1}{|p-p_{0}|} d^2{S}d^2{S}

Reply 3

ghostwalker

Original post by adie_raz

\frac{1}{2}N^2\frac{1}{16 \pi^2} \int_{S} \int_{S} \frac{1}{|p-p_{0}|} d^2{S}d^2{S}

I would have thought:

\displaystyle\frac{1}{2}N^2\frac{1}{16 \pi^2} \int_{S} \frac{1}{|p-p_{0}|}\;\mathrm{dS}

(edited 11 years ago)

Reply 4

adie_raz

Original post by ghostwalker

I would have thought:

\displaystyle\frac{1}{2}N^2\frac{1}{16 \pi^2} \int_{S} \frac{1}{|p-p_{0}|}\;\mathrm{dS}

I think it's because I have fixed the point

p_0

and when I do this I must multiply by the surface area of the sphere. Which sorts my original issue. I shall look into notation and post again when I have an answer. I would agree with you though, it seems I wasn't understanding the notation when I took the notes back in November!

Reply 5

adie_raz

Ok so I think it goes as follows:

\frac{1}{2}N^2\frac{1}{16\pi^2} \int_{S} \int_{S'}\frac{1}{|p-p_{0}|} dS'dS

= \frac{1}{2}N^2\frac{1}{16\pi^2} \int_{S} \int_{0}^{2\pi}\int_{0}^{\pi} \frac{r^2sin\theta}{\sqrt{2-2cos\theta}} d\theta d\phi dS

= \frac{1}{2}N^2\frac{4\pi}{16\pi^2}\int_{S} dS

= \frac{1}{2}N^2\frac{4\pi}{16\pi^2} \int_{0}^{2\pi}\int_{0}^{\pi} r^2sin\theta d\theta d\phi

= \frac{1}{2}N^2\frac{4\pi \times 4\pi}{16\pi^2}

= \frac{1}{2}N^2

(edited 11 years ago)

Reply 6

ghostwalker

Original post by adie_raz

...

One thing I would query, is why you have the 16/pi^2 to start with - but I'm not up on the physics behind this.

Reply 7

adie_raz

Original post by ghostwalker

One thing I would query, is why you have the 16/pi^2 to start with - but I'm not up on the physics behind this.

Sorry, I have found that is the charge density :smile:

I am not up on the physics of it either to be honest, I need to look into it more, but at least I know where I was going wrong now, thanks for the help, always useful to bounce ideas around!