# Electric field inside a uniformly polarised cylinder

• niko_.97
In summary, the conversation discusses problem 4.13 from Griffiths (edition 3) which asks for the electric field inside a long cylinder with uniform polarization perpendicular to its axis. The conversation explores two different methods for finding the electric field, one involving finding bound charges and the other using Gauss's law. There is some confusion about the results obtained using these methods and the conversation also touches upon finding the electric potential using an integral formula. Ultimately, it is noted that the integral solution for the electric potential is not a simple one and may require solving a boundary value problem.
niko_.97
Moved from a technical forum, so homework template missing
This is problem 4.13 from Griffiths (edition 3).
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?

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.

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.

niko_.97 said:
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?

niko_.97 said:
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?

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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!

niko_.97 said:
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.

## What is an electric field inside a uniformly polarised cylinder?

An electric field inside a uniformly polarised cylinder is a type of electric field that is created when the cylinder is placed in an external electric field. The polarisation of the cylinder causes a redistribution of charges, resulting in an electric field inside the cylinder.

## How is the electric field inside a uniformly polarised cylinder calculated?

The electric field inside a uniformly polarised cylinder can be calculated using the formula E = σ/ε0, where σ is the surface charge density and ε0 is the permittivity of free space. This formula assumes that the cylinder is infinitely long and has a constant polarisation throughout its length.

## What is the direction of the electric field inside a uniformly polarised cylinder?

The direction of the electric field inside a uniformly polarised cylinder is parallel to the axis of the cylinder and points towards the positive end of the cylinder. This is due to the alignment of the polarised molecules in the cylinder.

## How does the electric field inside a uniformly polarised cylinder differ from the electric field inside a non-polarised cylinder?

The electric field inside a uniformly polarised cylinder is stronger than the electric field inside a non-polarised cylinder. This is because the polarisation of the cylinder creates an additional electric field that adds to the external electric field.

## What are some real-world applications of the electric field inside a uniformly polarised cylinder?

The electric field inside a uniformly polarised cylinder has various applications, including in capacitors, where it helps to store electric charge. It is also used in devices such as electromagnets, particle accelerators, and MRI machines. In addition, it is crucial in understanding the behaviour of materials in electric fields, which has implications in fields such as materials science and engineering.

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