Electric field inside a uniformly polarised cylinder

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Moved from a technical forum, so homework template missing
This is problem 4.13 from Griffiths (edition 3).
The question asks:
A very long cylinder, of radius a, carries a uniform polarization P perpendicular
to its axis. Find the electric field inside the cylinder. [Careful: I said "uniform," not "radial"!]

I decided to try and find the bound charges. I am not really sure what he means exactly by uniform and perpendicular to the axis, so I just took it as being in the x hat direction. (i.e P=Px̄)
The bound volume charge is zero since the divergence of x hat is 0. Then the bound surface charge I got as being σb = Pcos(φ) for 0≤φ≤π and σb = -Pcos(φ) for π<φ<2π for the curved side of the cylinder and then 0 for both the top and bottom face. If you integrate this to find the potential you get zero (since the integral of cosφ from 0 to π is 0). Obviously then the electric field is also 0.

However, the solutions do it a different method.
Think of it as two cylinders of opposite uniform charge density ±ρ. Inside, the field at a distance s from
the axis of a uniformly charge cylinder is given by Gauss’s law: E2πsl = (1/ε0)ρπs2l => E = (ρ/2ε0)s. For
two such cylinders, one plus and one minus, the net field (inside) is E = E+ + E- = (ρ/2ε0) (s+ s). But
s+ s= −d, so E = −ρd/(2ε0), where d is the vector from the negative axis to positive axis. In this case
the total dipole moment of a chunk of length l is Pπa2l=ρπa2ld. So => ρd = P, and E =−P/(2ε0), for s < a.

I think I understand the solution. although why d=/=0 isn't exactly clear to me.
But my main query is why my solution gives you a different answer?
 

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  • #2
Charles Link
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In the integral that you did to find the potential, is ## r ## the same everywhere? That gets the result that the potential is zero at the center of the cylinder, but do you get that result at all locations? ## \\ ## See also https://www.physicsforums.com/threa...ormly-polarized-cylinder.941830/#post-5956930 where someone else had a question, just the other day, about Griffith's solution to this problem. ## \\ ## The same technique could also be used to find the electric field outside the cylinder, and could also be used to solve the problem of the electric field for a uniformly polarized dielectric sphere.
 
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In the integral that you did to find the potential, is ## r ## the same everywhere? That gets the result that the potential is zero at the center of the cylinder, but do you get that result at all locations?
I kept r (the vector from the origin to the point I want to find the potential at), just as r. But since we are not integrating over r I don't think that would matter.
I also set the r' in the integral equal to R, since all the surface charge is at a distance R.

Thanks for the link, although on there you just talk about the solution he gives in the textbook.
 
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Charles Link
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I kept r (the vector from the origin to the point I want to find the potential at), just as r. But since we are not integrating over r I don't think that would matter.
I also set the r' in the integral equal to R, since all the surface charge is at a distance R.
The formula for the electric potential at location ## \vec{r} ## reads ## V(\vec{r})=\int \frac{\rho(\vec{r}')}{4 \pi \epsilon_o |\vec{r}-\vec{r}'|} \, d^3r' ##, with a simple extension to the case where the charge density is a surface charge density. ## \\ ## ## \rho(\vec{r}') ## is the electrical charge density at location ## \vec{r}' ##. ## \\ ## The integral is only zero at location ## \vec{r}=(0,0,0) ##. [(x,y,z) coordinates here so that there is no ambiguity]. ## \\ ## Incidentally, the integral for ## V(\vec{r}) ## for this geometry is not a simple one. Griffith's has a somewhat unique and clever approach to finding the electric field for this case. The alternative method of finding the electric field usually proceeds with writing out the Legendre polynomial type solutions, (both inside and outside the cylinder), to Laplace's equation, and solving the boundary value problem. This process is not a simple one, but is perhaps easier than trying to evaluate the integral for the potential as a function of ## \vec{r} ##.
 
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The formula for the electric potential at location ## \vec{r} ## reads ## V(\vec{r})=\int \frac{\rho(\vec{r}')}{4 \pi \epsilon_o |\vec{r}-\vec{r}'|} \, d^3r' ##, with a simple extension to the case where the charge density is a surface charge density. ## \\ ## ## \rho(\vec{r}') ## is the electrical charge density at location ## \vec{r}' ##. ## \\ ## The integral is only zero at location ## \vec{r}=(0,0,0) ##. [(x,y,z) coordinates here so that there is no ambiguity].
That's what I tried to use (except I didn't set r to 0 from the start though it shouldn't matter surely?). The integral I ended up getting (just for the interval 0 to pi for phi) was:

V(r)=PR2/(4πε0)∫0π0π (cos(φ)sin(θ))/√(r2-R2-2rRcos(θ)) dθdφ (sorry about how it looks, still trying to get used to how to use the forum)
Is this right? If so, then V should be 0?
 
  • #6
Charles Link
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That's what I tried to use (except I didn't set r to 0 from the start though it shouldn't matter surely?). The integral I ended up getting (just for the interval 0 to pi for phi) was:

V(r)=PR2/(4πε0)∫0π0π (cos(φ)sin(θ))/√(r2-R2-2rRcos(θ)) dθdφ (sorry about how it looks, still trying to get used to how to use the forum)
Is this right? If so, then V should be 0?
Please see the additions to my previous post which I just added a minute ago. I think those additions might help answer your question. And to answer the latest question here, no, the result for the integral is not zero. You need to work in cylindrical coordinates, (and not spherical coordinates), and you won't have a mixture of ## \theta ## and ## \phi ##. Instead, the integral will be complicated with a ## z ##, and the location ## \phi ## will be in there, along with ## \phi' ## over which you are integrating. ## \\ ## Additional note: This integral in cylindrical coordinates is difficult enough that I actually have never seen any textbook that evaluates it. The standard approach is evaluating the potential using Legendre polynomials, but Griffith's solution is a very clever shortcut to the answer. ## \\ ## Additional item: For the problem of a dielectric sphere in a uniform field, (spherical coordinates), the standard approach is also a Legendre polynomial solution. Griffith's technique, which is kind of unique, gets you a much quicker solution. It was only quite recently that I managed to solve the integral for the potential in the case of a uniformly polarized sphere =see the discussion following this Insights article where I posted the solution: https://www.physicsforums.com/insights/homopolar-generator-analytical-example/ (To access the discussion, click on the part at the bottom "Discuss in the Community", and see post 9 of the discussion for the evaluation of the integral for the spherical case).
 
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  • #7
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Please see the additions to my previous post which I just added a minute ago. I think those additions might help answer your question. And to answer the latest question here, no, the result for the integral is not zero. You need to work in cylindrical coordinates, (and not spherical coordinates), and you won't have a mixture of ## \theta ## and ## \phi ##. Instead, the integral will be complicated with a ## z ##, and the location ## \phi ## will be in there, along with ## \phi' ## over which you are integrating. ## \\ ##
Ah OK, I'm still not 100% sure as to why the integral doesn't work, but I guess I've set the problem up wrong. Theoretically, there's no reason why you shouldn't be able to do the integral in spherical coordinates (although it'll be a lot more complicated), so I think if I do I should get some new function of phi in the integral to insure it's not 0. I understand Griffith's solution, I just wouldn't have thought to do this myself.

Thanks, for the help!
 
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  • #8
Charles Link
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Ah OK, I'm still not 100% sure as to why the integral doesn't work, but I guess I've set the problem up wrong. Theoretically, there's no reason why you shouldn't be able to do the integral in spherical coordinates (although it'll be a lot more complicated), so I think if I do I should get some new function of phi in the integral to insure it's not 0. I understand Griffith's solution, I just wouldn't have thought to do this myself.

Thanks, for the help!
I should point out that a cylinder in spherical coordinates does not have ## r=a ##. Unless the ## \phi ## integral is zero, (and it isn't), when you try to integrate over the surface area of the cylinder in spherical coordinates, you are going to have some very clumsy expressions, particularly for ## r ## as a function of ## \theta ## to keep the integration on the surface. ## \\ ## In cylindrical coordinates, it is quite easy to stay on the surface of the cylinder in doing the integration over the surface. The one problem is that the expression for the potential, basically the term ## \frac{1}{|\vec{r}-\vec{r}'|} ##, makes the integral quite complex and very difficult to evaluate.
 

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