Deriving potential distribution between concentric spherical electrode

[will6459]

Homework Statement

For some work I am doing I wish to be able to define the potential distribution as a function of the radius (ρ) between two concentric electrodes.

Homework Equations

One solution (from reliable literature) defines the varying radial potential as:

V(ρ)=2V0(ρ0/ρ -1)

Where V0 and ρ0 are the potential and radius of the optic axis respectively.

The Attempt at a Solution

Taking the laplace equation in spherical coordinates I find the potential is given by:

\frac{∂^{2}V}{∂ρ^{2}}=\frac{1}{ρ^{2}}\frac{∂}{∂ρ}(ρ^{2} \frac{∂V}{∂ρ})=0

This assumes there is no variance in potential in θ or \varphi which I'm confident is correct providing there is a uniform surface charge distribution across both electrodes. However the steps after this become pretty complex involving Legendre polynomials. I'm certainly no where near deriving the above equation.

Any help or suggestions anyone may have would be very welcome!

 

[Stealth95]

Hello.
It is possible to solve this equation without using Legendre polynomials. Just remember that:
\displaystyle{\frac{\partial}{\partial \rho }\left(\rho ^2\frac{\partial V}{\partial \rho } \right)=0\Rightarrow \rho ^2\frac{dV}{d\rho }=C}
which is easy to solve. You have to find two constants using two known values of the potential. I am not sure which are these values at your exercise, but I can find a solution which is the same as yours.

 

[will6459]

Thats a great help thank you.

So I rerrange and integrate to give:

\frac{dV}{dρ}=\frac{C}{ρ^{2}}

\int\frac{C}{ρ^{2}}dρ=\frac{-C}{ρ}+D

So:

V(ρ)=\frac{-C}{ρ}+D

I know that at the optic axis (ρ0), V(ρ)=V0 which leaves me with:

V_{0}=\frac{-C}{ρ_{0}}+D

I'm struggling to work out what the other boundary may have been used for the quoted solution in my previous post. If I examine ρ=∞ where V=0, then I get:

0=0+D

Thus V(ρ) is simply:

V(ρ)=\frac{V_{0}ρ_{0}}{ρ}

But this clearly doesn't match my quoted solution:

V(ρ)=2V_{0}(\frac{ρ_{0}}{ρ}-1)

Thanks for the reply Stealth, very helpful!

P.S. I'm unsure as to where the 1/ρ^{2} has gone in your previous post Stealth