# Evaluation of potential, problem reduced to an integral

• fluidistic
In summary, The conversation discusses using Green's function to calculate the potential and electric field due to a spherical shell with uniform charge density. The attempt at a solution involves calculating the potential function along the z-axis and using a change of variable, but there is an error in the final integral. The conversation also mentions the possibility of adding an arbitrary constant to the potential and making it continuous across the sphere.
fluidistic
Gold Member

## Homework Statement

I must use Green's function $\Phi (x)=\int _{\mathbb{R}^3 } \frac{\rho (x')}{|x-x'|}d^3x'$ to calculate the potential and the electric field due to a spherical shell of radius R and unfiform charge density $\sigma _0$.

P.S.:I know I could use Dirac's delta to get the solution faster but I'm asked to do so in another exercise.

## Homework Equations

Already given but in my case it reduces to (I believe): $\int _S \frac{\sigma _0 r'^2 \sin \theta d\theta d\phi }{\sqrt{ r^2+r'^2-2rr' \cos \gamma }}$ where $\cos \gamma = \cos \theta \cos \theta '+ \sin \theta \sin \theta ' \cos (\varphi ' - \varphi )$.

## The Attempt at a Solution

I assume the potential function phi to be radially symmetrical, in other words I can calculate $\Phi (r)$ and that would be the solution for whatever phi and theta.
Let's say I calculate Phi along the z-axis so that theta=0. This gives $\cos \gamma = \cos \theta '$.
Therefore $\Phi (r)=\int _0^{2\pi } \int _0^{\pi } \frac{r'^2 \sin \theta ' d\theta ' d \varphi ' }{\sqrt {r^2+r'^2-2rr' \cos \theta ' } }$$=2 \pi \sigma _0 \int _0 ^{\pi }\frac{r'^2 \sin \theta ' d \theta ' }{\sqrt {r^2+r'^2 -2rr' \cos \theta ' } }$. Now I make the change of variable $x=\cos \theta ' \Rightarrow dx=-\sin \theta '$.
And thus $\Phi (r)=2\pi \sigma _0 \int _{-1}^1 \frac{r'^2 dx}{\sqrt {r^2+r'^2-2rr'x} }$. This is where I'm stuck. Basically r' is a constant... oh wait then I guess I should look into an integral table?
If you see any error feel free to help me.

Edit: The integral is worth $-\frac{\sqrt {r^2-2rr'x+r'^2 } }{rr'} \big | _{-1}^1$.
Giving me $\Phi (r)=\frac{ 2\pi \sigma _0}{rr'} (\sqrt {r^2+r'^2+2rr'}- \sqrt {r^2+r'^2-2rr'} )$. I'm sure there's an error somewhere, the square roots are "too ugly". By the way $r'=R=\text {constant}$.

Last edited:
Complete the square in the square root in your final step, the answer looks clean. The dimension of your answer is sigma/meter, while before you evaluate the integral it is sigma*meter, somewhere you missed some r^2.

sunjin09 said:
Complete the square in the square root in your final step, the answer looks clean. The dimension of your answer is sigma/meter, while before you evaluate the integral it is sigma*meter, somewhere you missed some r^2.

Oops you are right about the r'² term, I forgot it. Ok about completing the square...

I get $$\Phi (r)=4\pi \sigma _0 \frac{R^2}{r}$$ if $r>R$.
And $\Phi (r)=0$ if $r<R$ which, if I recall well is false. I think it should be constant rather than null.

remember potential is not a physical quantity, you can add arbitrary constant to it and still valid. In order to make your potential look nice, you can afford to make it continuous across the sphere.

sunjin09 said:
remember potential is not a physical quantity, you can add arbitrary constant to it and still valid. In order to make your potential look nice, you can afford to make it continuous across the sphere.

Thanks for helping me.
So here I could "define" $\Phi (r)=4 \pi \sigma _0 R$ when r<R. Is this right?

## 1. What is the purpose of evaluating potential and reducing a problem to an integral?

Evaluating potential and reducing a problem to an integral allows for a more efficient and accurate analysis of complex systems or phenomena. It helps to simplify the problem and find a single solution that can be easily applied.

## 2. How is potential evaluated in scientific research?

In scientific research, potential is often evaluated through mathematical models and equations. These models take into account various factors and variables to determine the potential outcome or solution to a problem.

## 3. Can potential be accurately predicted through evaluation and integration?

While evaluation and integration can provide a good estimate of potential, it is important to note that there are always uncertainties and limitations in scientific research. Therefore, the predicted potential may not always be 100% accurate.

## 4. Are there any limitations to evaluating potential and reducing a problem to an integral?

Yes, there can be limitations such as the complexity of the problem, the accuracy of the data and models used, and the assumptions made in the evaluation process. It is important to consider these limitations when interpreting the results.

## 5. How does evaluating potential and reducing a problem to an integral benefit scientific advancements?

By utilizing this method, scientists are able to better understand and predict the behavior of complex systems. This can lead to advancements in various fields such as engineering, physics, and biology, allowing for more efficient and effective solutions to real-world problems.

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