# Deriving potential distribution between concentric spherical electrode

• will6459
In summary, the conversation discusses finding the potential distribution between two concentric electrodes using the laplace equation in spherical coordinates. The potential is given by V(ρ)=2V0(ρ0/ρ -1), and it can also be solved without using Legendre polynomials by using the equation \partial/\partial\rho(\rho^2\partial V/\partial\rho)=0. The constant C is found by using two known values of the potential, and the final solution is V(ρ)=-C/ρ+D. However, this solution does not match the quoted solution of V(ρ)=2V0(ρ0/ρ-1).
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!

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.

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:

$V(ρ)=2V_{0}(\frac{ρ_{0}}{ρ}-1)$

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

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## What is the purpose of deriving potential distribution between concentric spherical electrodes?

The purpose of deriving potential distribution between concentric spherical electrodes is to understand and predict the electric potential and field distribution between two or more spherical electrodes. This information is crucial in designing and analyzing various electrical systems and devices, such as capacitors, batteries, and sensors.

## What are the assumptions made when deriving potential distribution between concentric spherical electrodes?

The main assumptions made when deriving potential distribution between concentric spherical electrodes are:

• The electrodes are perfectly spherical.
• The electrodes are concentric, meaning they share the same center point.
• The materials of the electrodes are homogeneous and isotropic.
• The medium between the electrodes is a perfect insulator.
• The electric potential is constant on the surface of each electrode.

## How is the potential distribution between concentric spherical electrodes calculated?

The potential distribution between concentric spherical electrodes is calculated using the Laplace equation, which describes the relationship between the electric potential and the charge distribution in a system. The equation is solved using mathematical techniques, such as separation of variables and boundary conditions, to obtain the potential function.

## How does the distance between the electrodes affect the potential distribution?

The distance between the concentric spherical electrodes has a direct impact on the potential distribution. As the distance increases, the potential decreases, following an inverse square law. This means that the potential decreases exponentially as the distance between the electrodes increases.

## What are the practical applications of deriving potential distribution between concentric spherical electrodes?

The potential distribution between concentric spherical electrodes has many practical applications in various fields, including:

• In the design and analysis of capacitors, which use concentric spherical electrodes to store electric charge.
• In the development of biomedical devices, such as pacemakers and defibrillators, which use concentric electrodes to deliver electric pulses to the heart.
• In the study of Earth's electric field and atmospheric electricity, as the Earth can be approximated as a concentric spherical electrode system.

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