Electromagnetism, potential

[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

[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.

[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

[ghosts_cloak]

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

[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}

[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

[maverick6664]

oh yeah, it says calculate potential first! lol.

[maverick6664]

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

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.

[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...