# Electromagnetism, potential

1. Jan 26, 2006

### ghosts_cloak

http://www.zen96175.zen.co.uk/question.GIF

Hi everyone :-) This is my first post here!
I have a question regarding electrostatics, in the picture above. I am not looking for the solution at all as this work will be assesed, Im just hoping that someone can verify/ poke me in the right direction.

First of all potential is q/(k*x) in x-hat direction?
Integrate that over 2*Pi*a to get potential of a hoop.
Using cosine rule to express x in terms of theta, then integrate over whole hoop from 0 to Pi..
Thats what I have been trying, and Im not having any joy. Any small pointers would be most appreciated!

I hope that was okay for a first post!

~Gaz

2. Jan 26, 2006

### maverick6664

You are almost correct. Wrong point is all potential is $$4\pi a^2 k \sigma / r = kq/r (or \frac q {4\pi \epsilon_0 r} )$$

First think of the ring portion on the sphere between $$\theta$$ and $$\theta + d \theta$$ (area is $$2\pi a^2 \sin \theta d \theta$$), and calculate the electric field the ring creates at P (note you only need to calculate horizontal factor ignoring other directions.) Then integrate it over $$0 \leqq \theta \leqq \pi$$.

That's all and I'm also calculating it.

Last edited: Jan 26, 2006
3. Jan 26, 2006

### ghosts_cloak

Hi, thanks for the reply :-)
I am a little mystified by the area you mention... in what orientation are you taking this ring portion?
Maybe it will come to me, I am working on the problem now but at the moment I cant see it....
Thanks!

~Gaz

4. Jan 26, 2006

### ghosts_cloak

Actually, I see where you get the area from now :-) Ill keep at it..
~Gaz

5. Jan 26, 2006

### maverick6664

under construction :)

ok. I've done all the calculations.

First, the electric charge in the ring on the sphere between $$\theta and \theta + d \theta$$ is $$\2pi a^2 \sigma$$. So the electric field at P this ring contributes is $$dE = \frac {2\pi a^2 \sigma \sin \theta d \theta} {x^2} \cos \angle QPO$$ (Q is the point on the sphere at $$\cos \theta$$. You can take into account only the horizontal component of the electric field, so you need to multiply $$\cos \angle QPO$$. So consider $$x^2 = a^2 + r^2 - 2ar \cos \theta$$ because of law of cosines,

$$dE = \frac {2\pi a^2 \sigma \sin \theta } {x^2} \cdot \frac {r-a \cos \theta} { x} = 2 \pi a^2 \sigma (r^2+a^2 - 2ar \cos \theta)^{-3/2} (r-a \cos \theta) \sin \theta d \theta$$

We need to integrate $$0 \leqq \theta \leqq \pi$$, but substitute $$u = \cos \theta,$$ so $$d \theta = \frac {d \theta}{du} du = - \frac 1 {\sin \theta} du$$.

So we integrate:

$$dE = 2 \pi a^2 \sigma (r^2+a^2 -2aru)^{-3/2}(r-au)du$$ over $$-1 \leqq u \leqq 1$$ (note sign is reversed.)

Using partial integral,

$$E = 2\pi a^2 \sigma (\frac 1 {ar} (r^2+a^2 -2aru)^{-1/2} (r-au) \vert_{-1}^1 + \frac 1 {ar} \int_{-1}^1 a(r^2+a^2-2aru)^{-1/2} du)$$

$$= 2\pi a^2 \sigma (\frac 1{ar} (\frac {|r-a|}{|r-a|} - \frac {|r+a|} {|r+a| } ) + \frac 1{ar} (-\frac {2a}{2ar} (r^2+a^2-2aru)^{1/2} \vert_{-1}^1)$$

$$= 2\pi a^2 \sigma(\frac 1{ar} \frac 1 r(-(|r-a|-|r+a|))$$
$$= \frac {4 \pi a^2 \sigma} {r^2}$$

Last edited: Jan 26, 2006
6. Jan 26, 2006

### ghosts_cloak

Hiya, thanks for the effort you have put in!
I think I have the solution now, I found the potential first and then the electric field from that, and everything drops out nicely and appears like a point charge as required. I will of course work through your solution also to compare, I thought mine was correct although its a lot shorter than yours...Ill have to check I havent missed anything!

~Gaz

~Gaz

7. Jan 26, 2006

### maverick6664

oh yeah, it says calculate potential first! lol.

8. Jan 26, 2006

### maverick6664

Yeah! If you calculate potential, you don't have to consider $$\cos \angle QPO$$, so it's much shorter and easier! This cosine part makes the calculation complicated, because of this part, partial integral is necessary.

9. Jan 27, 2006

### maverick6664

My calculation for potential is (as the question requests

$$d \phi = \frac {2\pi a^2 \sigma \sin \theta} x d \theta$$

so it follows that:

$$\phi = \int_0^\pi \frac {2 \pi a^2 \sigma \sin \theta} x d \theta$$

$$= \int_0^\pi 2\pi a^2 \sigma (r^2+a^2 -2ar \cos \theta)^{-1/2} \sin \theta d \theta$$

$$= \int_{-1}^1 2 \pi a^2 \sigma (r^2+a^2 - 2aru)^{-1/2} du = \frac {4 \pi \sigma a^2} r$$

actually much easier...

Last edited: Jan 27, 2006