# Electric potential using Green function, particular set up

1. Feb 23, 2013

### fluidistic

1. The problem statement, all variables and given/known data
I would like to calculate the electric potential in all the space in the following set up:
Conductor sphere of radius a whose surface is kept at a zero potential. 1 point charge $q_1$ at distance $d_1$ from the center of the sphere. 1 point charge $q_2$ at a distance $d_2$ from the center of the sphere.
There's the relation $d_1>d_2>a$
The set up is such that if you use a Cartesian coordinate system with x and y-axis centered in the center of the sphere, q1 is at $(0,d_1)$ and q2 is at $(d_2,0)$.

2. Relevant equations
Green function for a sphere: $G(x,x')=\frac{1}{|x-x'|}-\frac{a}{|x'|x-\frac{a^2x'}{|x'|^2}}$.
Then, the potential is given by $\Phi (x)= \int _ \Omega G(x,x')\rho (x')d^3x'+\frac{1}{4\pi} \oint _{\partial \Omega} \frac{\partial G (x,x')}{\partial n' }V(x')da'$.

3. The attempt at a solution
I'll use spherical coordinates $(r, \theta , \phi )$.
I want to write the Green function in spherical coordinate and also the charges distributions.
First I notice that $V(x')=0$, because that's the potential over the sphere. So that simplifies greatly the potential expression.
I'm left with $\Phi (x)= \int _ \Omega G(x,x')\rho (x')d^3x'$ where I must write all under spherical coordinates. I think I've $G(x,x')$ under spherical coordinate on my sheet.
Now I'm unsure about $\rho (x')$. The first charge distribution would be... $\rho _1 (\vec x )=q_1 \frac{\delta (\theta ) }{r} \delta (r-d_1 ) \frac{\delta (\phi)}{r \sin \theta }$. I'm unsure mostly about the Dirac's delta of the phi coordinate. Because that angle is not definied for spherical coordinate system, here I took it as the charge is at phi=0, but it could have been any other number, hence my doubt.
For the 2nd charge, $\rho _2 (\vec x ) =q_2 \frac{\delta (\theta - \frac{\pi}{2}) \delta (r-d_2) \delta (\phi) }{r^2 \sin \theta }$.
Here I'm unsure again. I'm doubting whether the surface of the sphere gets polarized, in which case I guess I'd have to include a $\rho _3 (\vec x )$ that contributes as a source. But since V=0, I'm not really sure that it gets polarized though I believe it really does.
This means I'd have to also calculate $\sigma _{\text{induced }}$ at the surface of the sphere, but I don't even have the potential in all the space nor the electric field.
So I don't know how to continue further. Any help is appreciated.

P.S.:I'm using cgs system of units where the Coulomb constant is worth k=1 while Jackson's and Griffiths use SI units where $k=\frac{1}{4\pi \varepsilon _0}$.

Last edited: Feb 23, 2013
2. Feb 24, 2013

### Mute

Why is $\phi$ not defined in the spherical coordinate system? In spherical coordinates, $x = r\cos\phi \sin\theta$ and $y = r \sin\phi \sin\theta$ using your angle convention, I believe, so if the first charge is located at $x = d_1$ and $y=0$. $\sin\theta$ can't be zero, so that implies $\phi = 0$. Presumably, we want to align our coordinate system such that the two charges lie in a plane, so we choose for the two charges to lie in the plane $z = r\cos\theta = 0$, which gives $\theta = \pi/2$. (Since $\theta$ is measured from the z-axis).

Note: if your convention for the spherical coordinates has $\phi$ and $\theta$ switched relative to mine, then I believe your volume element used with the dirac delta function is incorrect. Given $x = r\cos\phi \sin\theta$, $y = r\sin\phi \sin\theta$ and $z = r\cos\theta$ as the definition of the spherical coordinates, the volume element is $dV = dr~d\phi~d\theta~r^2 \sin\theta$. So, make sure your coordinate definitions and your volume element are consistent.

It's a bit late here, and it's been a while since I did an EM problem like this, so I'm a bit hazy on what exactly you should do in this situation. My first instinct is to just solve the problem outside of the sphere, since the surface of the sphere is set to V=0. For the inside of the sphere, I have to ask: is this an actual SPHERE in the precise mathematical sense - i.e., a hollow ball, where "sphere" refers to the actual surface - or is it really a ball full of material, and only the surface of the material is set to V = 0? This would affect the density you have to use inside the sphere, I think. Beyond this, it's too late for me to offer further suggestions - hopefully someone in just getting up in a different timezone can take it from here.

3. Feb 24, 2013

### PhysicsGente

Why don't you use images. It'll be a lot easier ;). If not, I'd suggest reading Schwinger (ch. 19 I think). He goes goes into a lot of detail for this kind of calculations.

4. Feb 24, 2013

### fluidistic

First of all, thank you very much for the help to both.
and $z= r\cos \theta$. In my draft the charge q1 lies over the z-axis where the coordinate phi is not defined. It could be anything and for the Dirac's delta I assumed phi=0 but I just realized I believe I can't just assume it's 0, I'll explain this below.
This is exactly what I've done on my draft (I mean they lie on a plane). I chose q1 to be at $(0,0,d_1)$ and q2 at $(d_2,0,0)$ (in Cartesian coordinates) so both charges lie in the x-z plane. Sorry for not having precised this before.
So my doubt is, how do I specify the charge distribution of q1 in spherical coordinates?
I used the same convention as you and my volume element is exactly the same as yours :).

No problem! I appreciate your help.
Well I don't remember. The thing is, it's a conductor and an electrostatics problem so all the free charges are on the surface of the sphere. Since the electric field inside that sphere is 0, it means the potential is constant inside the sphere. And since the potential is continuous and equal to 0 at the surface of the sphere, then the potential must be worth 0 inside the sphere too. I believe there's no difference if the sphere was hollow or filled with a perfect conductor.
I'm only left to calculate the potential outside the sphere indeed.

The reason I said I can't simply assume $phi=0$ to describe $\rho _1 (\vec x )$, it's because in the formula $\Phi (x)= \int _ \Omega G(x,x')\rho (x')d^3x'$ where $\rho (x')=\rho _1 (\vec x_1 ) + \rho _2 (\vec x_2)+\rho _{\text{induced charge over the sphere?}}$, G(x,x') does depend on phi and so the Dirac's delta I have if I chose phi=0 is $\delta (\phi)$ and clearly if I had chosen for instance phi=100, I'd have $\delta (100- \phi)$ which would modify the value of the potential because of the Green function. So I just don't know which phi is "correct". Which phi gives the potential.

Good idea, I could solve the problem via both ways. I'm just unsure at first glance where to put the equivalent charge inside the sphere. It's not really obvious to me on what axis I must place it.

5. Feb 24, 2013

### PhysicsGente

Ok. Lets suppose you have only one charge $q$ in the y-axis at a distance $y$ from the center of the sphere. Place a charge $-q'$ on the y-axis at a distance $y'$ from the center of the sphere (inside the sphere).

Now, the potential is given by $\phi = \frac{q}{|\vec{r}-\vec{r_0}|} - \frac{q'}{|\vec{r}-\vec{r'_0}|}$.

Then, at the surface of the sphere we have that the potential is zero so $\phi = \frac{q}{|a\hat{y}-y\hat{y}|} - \frac{q'}{|a\hat{y}-y'\hat{y}|} = \frac{q}{a|\hat{y}-\frac{y}{a}\hat{y}|} - \frac{q'}{y'|\frac{a}{y'}\hat{y}-\hat{y}|} = 0$.

So $\frac{q}{a}=\frac{q'}{y'}$ and $\frac{y}{a}=\frac{a}{y'}$.

Those are the relations you need.

6. Feb 24, 2013

### fluidistic

Thanks again. This seems like the problem of a single charge outside a sphere, unlike the problem of this thread.
However I've sketched the situation, I got that the 2 unknowns are worth $y'=\frac{a^2}{y}$ and $q'=\frac{aq}{y}$.
So now I must repeat this process with a point charge q'' situated somewhere over the line between y' and q2? Where q2 is at $(d_2,0)$ in Cartesian coordinates.
In other words, how does the common method of images helps me for this problem?

7. Feb 24, 2013

### PhysicsGente

Your problem involves two charges, one in the y-axis and one in the x-axis. You already calculated what your image charge should be in the y-axis. Just do the same for the x-axis so that your potential at the surface of the sphere is zero. Then, all that is left is to rewrite the potential at an arbitrary point due to the four charges.

8. Feb 24, 2013

### fluidistic

I see, that's very bright.
I get $\Phi (\vec r ) = \frac{1}{4\pi \varepsilon _0 } \left [ \frac{q_1}{|\vec r - d_1 \hat y |} - \frac{q_1a}{d_1 |\vec r - \frac{a^2 \hat y }{d_1}|} + \frac{q_2}{|\vec r - d_2 \hat x |} - \frac{q_2a}{d_2 |\vec r - \frac{a^2 \hat x }{d_2}|} \right ]$.
Now I must rewrite $\vec r$, $\hat x$ and $\hat y$ in spherical coordinates. Or maybe Cartesian coordinates is easier after all!
By the way, that potential would be the potential outside the sphere right? But inside the sphere $\Phi =0$, right?
Thanks a lot for all!

9. Feb 24, 2013

### PhysicsGente

The electric field is zero within the conductor. So the potential is either zero or a constant. I guess it'll be zero then.

10. Feb 24, 2013

### fluidistic

Here's my reasoning: the potential must be continuous, else E "blows up" since $\vec E = - \vec \nabla \Phi$. Since the potential is worth 0 on the surface of the sphere and as you said it is constant inside the conductor, there's no other choice than to be equal to 0 inside the sphere.