Potential due to two oppositely charged half spherical shells

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Homework Statement


Consider the spherical shell of radius R shown in the figure characterized by a hemisphere of surface charge density [itex]\sigma[/itex] and total charge q in the upper half plane and a hemisphere of surface charge density -[itex]\sigma[/itex] and total charge -q in the lower half plane. Find the electric potential at:

(a) [itex]\vec{x}[/itex]=(7R,14R,0)
(b) and at [itex]\vec{x}[/itex] = [itex]\vec{0}[/itex]


Homework Equations



General form for a volume charge: V( [itex]\vec{r}[/itex]) = k [itex]\int\frac{\rho( \textbf{r'})}{<r>}d\tau'[/itex]
<r> is supposed to be a script r but I can't figure out how to make that worth in latex, sorry if that's confusing and I am open to suggestions!

The Attempt at a Solution


I believe (b) to be zero by observation. I am having trouble with (a).

-My first assumption is that the positive half is 'above' the x,y plane and the negative half is 'below' the x,y plane. It isn't made clear but it is not arbitrary considering I am given coordinates to analyze on the x,y plane. There is not a coordinate system or set of axis drawn in the diagram. Also, this is a little confusing because the point of interest is given as [itex]\vec{x}[/itex] and the convention that follows is a vector of magnitude x in the direction of x hat. He has stressed such conventions in other problems.

-I looked for similar examples to extrapolate from but the ones I find explicitly place the point of interest on the axis of symmetry (positive z, usually).

-If my first assumption is correct then the point of interest is on the x,y plane but all that really matters is the magnitude of the vector given (distance from origin) because the charge distribution is symmetric about the z axis. In other words, it doesn't matter if I rotate around the z in the x,y plane but does matter how far away I am. The x and y contributions of the positive cancel those of the negative and we are left with contributions in the z direction.

I tried to use this:

V( [itex]\vec{r}[/itex]) = k [itex]\int\frac{\rho( \textbf{r'})}{<r>}d\tau'[/itex]

to brute force integrate the charges separately and utilize superposition to add them together. I set it up like you would for a sphere (several examples available) but only integrated from pi/2 to 0 (starting on the x, y plane and rotating up to the z axis) so I get a half sphere instead. I have a result but I really can't tell if it makes any sense. I will happily post more on this if this route is reasonably suited for this type of problem.

To further complicate matters we did just learn about multipole expansions today but this problem doesn't ask for an approximation of the potential at large r. So I have to assume my prof wants an exact answer unless an exact answer is unreasonably difficult to come by.

Sorry for the wall of text but I am struggling with my intuition in this course and hope to be corrected where it fails. I am open to any suggestions. Thank you for your help!
 
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Answers and Replies

  • #2
Simon Bridge
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I believe (b) to be zero by observation. I am having trouble with (a).
What is the "by observation" for (b) being zero?

Note <r> could mean a 1D vector with r as it's only element, or the expectation value of r
magnitude of the vector ##\vec r## would be written ##|\vec r |## or just ##r##.

I think your professor is after an exact answer too.
Your initial approach taking rings of the sphere is OK - except that the position of ##\vec x## is in the x-y plane - so the symmetry is different.
It will help to sketch the situation - take an appropriate cross-section perhaps.

You will be doing a surface integral.
You want to work out the potential ##dV## at position ##\vec x## due to the charge element at some point on the sphere.
##dq = \sigma dS## where dS is the surface element at some position on the sphere.
What ##dS## and ##\vec r## ends up as depends on your choice of coordinate system.
 
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What is the "by observation" for (b) being zero?
Thanks for asking because as I began to type out an explanation I realized it wasn't that simple. I will have to re-consider.

Note <r> could mean a 1D vector with r as it's only element, or the expectation value of r
magnitude of the vector r⃗ \vec r would be written |r⃗ ||\vec r | or just rr.
I am most familiar with the expectation value interpretation of that notation and I realize that may be confusing. There are several r's in this example, however. Script r is used in Griffiths to note the distance from a segment of charge to the point of interest. R is the radius (constant), [itex]\vec{r}[/itex] as vector from origin to point of interest, then [itex]\vec{r}[/itex]' as vector from origin to segment of charge. I figured if I was explicit about that replacement it would serve its purpose as well as hint at the fact that I really still do want to know how to do a script r in latex here :) I was hoping someone would tell me how to do it and I could use script r's henceforth. Thanks for the recommendation.

I think your professor is after an exact answer too.
Your initial approach taking rings of the sphere is OK - except that the position of x⃗ \vec x is in the x-y plane - so the symmetry is different.
It will help to sketch the situation - take an appropriate cross-section perhaps.
I have a sketch but your suggestion brings us to my main issue with the original approach. I'm confused about how to re-arrange the spherical integral to deal with this modified symmetry. I made adjustments to the limits of the integral so that I at least end up with a half-sphere of charge but beyond that I'm stuck. Here is a summary of the 3d representation of my sketch:

ECZV7HE.jpg


For simplicity I left the point of interest on the y axis.

I also sketched a cross section using the z,x (or z,y) plane as the slicing plane. Which basically looks the same as above without the x axis and without the oval contained inside the circle which would otherwise express the 3d nature of the sphere.

You will be doing a surface integral.
You want to work out the potential dVdV at position x⃗ \vec x due to the charge element at some point on the sphere.
dq=σdSdq = \sigma dS where dS is the surface element at some position on the sphere.
What dSdS and r⃗ \vec r ends up as depends on your choice of coordinate system.
This is effectively the description of the integral I posted. Perhaps you are hinting at something here that I am not picking up, sorry! I suspect spherical coordinates would be the best bet given the spherical symmetry and can work out the surface elements for the integral once I understand conceptually how to deal with the symmetry.

To expound a bit on the above integral:

[itex]\rho(\vec{r})[/itex] is just an area element times our charge density. Since the density is constant it pops out and we're left with an area element inside the integral which in our case looks like [itex]R^2*sin(\theta)'d\theta'd\phi'[/itex] Then our script r is found using law of cosines.

Thanks for your input!
 
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  • #4
Simon Bridge
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For simplicity I left the point of interest on the y axis.
... that's OK because the symmetry supports this - you've effectively rotated the problem about the z axis. The new ##\vec x## will have the same magnitude but look like ##(0,x,0)##.

Your diagram has the point of interest - call it P to save typing - on the -y axis?
It is basically a good start, you can see a lot of the details from that.

This is effectively the description of the integral I posted. Perhaps you are hinting at something here that I am not picking up, sorry!
... I'm checking that I have understood you ;)
Your next task is to pick a coordinate system so you can make the dS and the r more specific.
You could start by finding the potential at P due to a semi-circular arc between y and y+dy and build from there?
 
  • #5
Orodruin
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Based on the points where you are asked for the potential I only have one word to help you along: symmetry.
 
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Sorry, I didn't realize you would reply so quickly. I made a few edits in the time that you were replying. I will add what I edited here so you don't have to re-read that post:

To expound a bit on the above integral:
I believe spherical would be the best given that we're basically integrating over the surface of the sphere.

[itex]\rho(\vec{r})[/itex] is just an area element times our charge density. Since the density is constant it pops out and we're left with an area element inside the integral which in our case looks like [itex]R^2*sin(\theta)'d\theta'd\phi'[/itex] Then our script r is found using law of cosines.

You could start by finding the potential at P due to a semi-circular arc between y and y+dy and build from there?
So basically slicing this thing using the z,x plane from y=-R to R?
 
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Based on the points where you are asked for the potential I only have one word to help you along: symmetry.
It's clearly symmetric about z and mirrored on the xy plane. Looking for symmetry is always my first approach in this class. I am clearly missing the connection between symmetry and its manifestation in an integral, however. This is conceptually difficult for me, sorry!

edit: Are you on the same page as Simon Bridge in looking at this system of charges from, say, the y-axis and taking slices which look like rings of charge?
 
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Orodruin
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It's clearly symmetric about z and mirrored on the xy plane. Looking for symmetry is always my first approach in this class. I am clearly missing the connection between symmetry and its manifestation in an integral, however. This is conceptually difficult for me, sorry!
The point is that it is anti symmetric about the z-axis as the lower shell has negative charge density.

Edit: and by "about the z-axis" I mean "under reflection in the xy plane". Early morning ...
 
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The point is that it is anti symmetric about the z-axis as the lower shell has negative charge density.
Symmetric as you rotate about z and anti symmetric about the xy plane. I mean, I realize that the problem has symmetries. It is blatantly spelled out in the problem and I've noted them several times in my replies. I am asking explicitly about how those symmetries manifest themselves in my solution. That's the very problem I am having.
 
  • #10
Orodruin
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If the charge distribution is anti symmetric, what can you say about the potential?
 
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If the charge distribution is anti symmetric, what can you say about the potential?
Well, in this case the net charge is zero. The potential is not zero, though, if that's what you are hinting at. If a positive test charge is placed on the x axis where x>R and we move this charge towards our origin the force on our test charge in the x direction would be 0 but the vertical components of the charge distributions would sum, not cancel. Perhaps you're hinting at something else though or maybe I'm flat out wrong?
 
  • #12
Orodruin
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A force on a particle means that the derivative of the potential is zero. The value of the potential can still be zero.

That the total charge is zero is a necessary consequence of the asymmetry of the charge distribution.
 
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I am starting to agree with you. I guess the force acting on this particle in the z direction doesn't count as work because it is perpendicular to the direction of motion... this is fundamental but not intuitive in this case from my perspective. That is the only force in the case of a dipole and this problem can be considered, via superposition, a series of dipoles. Not sure of a better way to word that but, yes, I guess I agree now that the potential at any point on the xy plane would be zero.

My question now is what you think about the way the question is worded. I listed it exactly as it is worded with all information given. It says nothing about the orientation of the half spheres besides noting one is above a plane and the other is below that plane. I guess I have to look at the problem assuming we can define this plane as the zx plane OR the xy plane, for example. We are still considering the potential on the xy plane in (a) though so the symmetry goes away if it is anti-symmetric about zx.
 
  • #14
Orodruin
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I would say the problem creator thought he/she was being explicit when saying "upper" but implicitly assuming the positive z direction as up. If you take away this implicit assumption the plane can be anything, not even necessarily a coordinate plane or passing through the origin (as another implicit assumption seems to be that the sphere is centered there) ... The answer is then pretty arbitrary.

A computation that will require actually doing some integrals is that for the potential on the z axis instead of in the xy plane, so this is my next challenge to you ;)
 
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  • #15
Orodruin
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I am starting to agree with you. I guess the force acting on this particle in the z direction doesn't count as work because it is perpendicular to the direction of motion... this is fundamental but not intuitive in this case from my perspective. That is the only force in the case of a dipole and this problem can be considered, via superposition, a series of dipoles. Not sure of a better way to word that but, yes, I guess I agree now that the potential at any point on the xy plane would be zero.
By "not counting as work" I assume you mean for a test particle that is moving along the xy plane because in the original problem there is nothing moving. That no work is performed on such a test particle is simply stating that it is moving along an equipotential surface (it could have a non-zero constant potential). What really gives you the zero is the anti symmetry (assuming you define the potential at infinity as zero).

And since you have now arrived at the correct answer: In order to make the connection between the anti symmetry and the potential being zero, consider the integral expression for the potential at a point ##(x,y,z)##:
$$
V(x,y,z) = k \int \frac{dq}{r} = k \int_{\mathbb R^3} \frac{\rho(x',y',z') dx' dy' dz'}{r}
$$
where ##r = r(x-x',y-y',z-z') = \sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}## and the charge density ##\rho## fulfils ##\rho(x',y',z') = - \rho(x',y',-z')##, i.e., is anti symmetric with respect to the xy plane. We rewrite the integral as
\begin{eqnarray}
V(x,y,z) &=& k \int_{\mathbb R^2} dx'dy' \int_{-\infty}^\infty dz' \frac{\rho(x',y',z') dx' dy' dz'}{r} = \{\zeta = -z'\} \\
&=& k \int_{\mathbb R^2} dx'dy' \int_{-\infty}^{\infty} d\zeta \frac{\rho(x',y',-\zeta) dx' dy' d\zeta}{r(x-x',y-y',z+\zeta)} \\
&=& k \int_{\mathbb R^2} dx'dy' \int_{-\infty}^{\infty} d\zeta \frac{-\rho(x',y',\zeta) dx' dy' d\zeta}{r(x-x',y-y',-z-\zeta)} \\
&=& -V(x,y,-z).
\end{eqnarray}
In the last step I have just used the anti symmetric property of ##\rho## and the symmetric property of ##r##. As a consequence ##V(x,y,0) = -V(x,y,-0) = 0##.
 
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