# Electric field due to a uniformly polarized cylinder

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1. Aug 8, 2017

### Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Uniform polarization : $σ_b = \vec P ⋅ \hat n$
Taking $\vec P = P \hat x ~and ~ \hat n = \hat s ~, ~ σ_b = P \cos \phi$

Let's take a strip of width R dΦ at an angle Φ. This strip will be equivalent to a rod of linear charge density λ = P R$\cos \phi$ , where R is the radius of the cylinder.
Taking the bottom of this strip as origin, electric field at Q due to this strip is
$\vec E_{strip} \left ( \vec r \right)= \frac {PR~\cos \phi d \phi} {2 \pi ε_0 s} \hat s$
Where I use s for curly r.

Now how to integrate for the whole cylinder.

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2. Aug 8, 2017

### BvU

Hi,

Can't make sense of your coordinates. Why not put the cylinder on the z axis, then it's complicated enough already.
Make a drawing in the $r,\phi$ plane
Griffiths uses $\hat s$ and you seem to be familiar with it. Please explain all used variables and symbols.
You might want to look at problem 3.25 too ...

Note that $\hat r$ and $\hat n$ don't coincide.

3. Aug 8, 2017

### Pushoam

I looked at it .
This leads me to Pro.3.23 which leads me to solve Laplace eqn. in cylindrical coordinates with z - symmetry.
I haven't done it. So, I will return after learning it.

Thank you.

4. Aug 9, 2017

### TSny

If you like, you can try to solve prob. 4.13 using the approach presented in example 4.3, which is just before the statement of prob. 4.13 in the third edition of Griffiths.

5. Aug 10, 2017

### Pushoam

Considering the polarized cylinder as superposition of two cylinders with opposite,equal, uniform charge densities in a way shown in the figure above.

Does this consideration mean the following ? :
In general, if I have an object with polarization $\vec P$ , then I have to take two objects (whose shape is similar to the system) with opposite, equal, and uniform charge densities and super-impose these two objects in such a way that the total dipole moment of this superimposed system is equal to the total dipole moment of original system. In this case, I can take the super- imposed system as equivalent to the original system for calculating electric field and potential.
But what does guarantee that this consideration is correct?

Electric field at M with position vector $\vec r$ is given by
$\vec E \left ( \vec r \right ) = \frac { Q\left ( \vec S_+ - \vec S_- \right ) }{ 2 \pi ε_0 a^2} = \frac { - \vec P } {2 ε_0}$, M is in the cylinder.

For M outside the cylinder,
$\vec E \left ( \vec r \right ) = \frac { Q\left (\frac{ \vec S_+} {S_+ ^2} - \frac { \vec S_- }{S_- ^2} \right ) }{ 2 \pi ε_0 l}$

$= \frac { Q \left ( { \vec S_+} {S_- ^2} - { \vec S_- }{S_+ ^2} \right ) }{ 2 \pi ε_0 l {S_+ ^2}{S_- ^2} }$

Now ,$\vec S_- = \vec S_+ + \vec d$
${S_- ^2} = {S_+ ^2} +d^2 + 2 \vec S_+ ⋅ \vec d ≈ {S_+ ^2} + 2 \vec S_+ ⋅ \vec d$
$\left ( { \vec S_+} {S_- ^2} - { \vec S_- }{S_+ ^2} \right ) = {S_+ ^2} \{ \vec S_+ - \vec S _- + 2 \left ( \vec d ⋅ \hat S_+ \right ) \hat S_+ \} \\= {S_+ ^2} \{ - \vec d + 2 \left ( \vec d ⋅ \hat S_+ \right ) \hat S_+ \}$

$\vec E \left ( \vec r \right ) = \frac { Q \{- \vec d + 2 \left ( \vec d ⋅ \hat S_+ \right ) \hat S_+ \} } { 2 \pi ε_0 l {S_- ^2} }$

For very small d, Taking approximation ( Is this approximation correct?) ${S_- ^2} ≈ s^2 ~, ~ \hat S_+ ≈ \hat s$, we get,

$\vec E \left ( \vec r \right ) = \frac { Q \{- \vec d + 2 \left ( \vec d ⋅ \hat s \right ) \hat s \} } { 2 \pi ε_0 l {s ^2} } \\= \frac { \{ -\vec p + 2 \left ( \vec p ⋅ \hat s \right ) \hat s \} } { 2 \pi ε_0 l {s ^2} }$
$\\ = \frac { a^2 \{2 \left ( \vec P ⋅ \hat s \right ) \hat s - \vec P \} } { 2 ε_0 {s ^2} }$

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6. Aug 10, 2017

### TSny

You can show that this consideration produces exactly the same distribution of surface charge as would exist on the uniformly polarized cylinder as long as you choose $\rho d = P$. Here $\rho$ is the magnitude of the volume charge density of one of the superimposed cylinders, $d$ is the displacement of the cylinders, and $P$ is the magnitude of the polarization of the polarized cylinder. This trick will work for arbitrary shapes as long as the polarization is uniform.

Looks good except that I think you left out the length $l$ of the cylinder in the middle expression. It might have been nice to write the middle expression in terms of $\rho$ instead of the total charge $Q$. Then you can make use of the relation $\rho d = P$ to get the expression on the right.

For outside the cylinder, your work looks very good to me.
Yes, the approximation is correct. You are looking for $\vec E$ to first order in $d$. So, you can see that the expression for $\vec E$ above will remain correct to first order in $d$ if you use "zero-order" approximations for $S_-$ and $\hat S_+$.

7. Aug 11, 2017

### Pushoam

I have to show here that the field due to the super-imposed system is equal to the field due to a cylinder with surface charge density $\vec P ⋅ \hat n$.
I have calculated field due to the super-imposed system. Now to calculate field due to the cylinder with surface charge density $\vec P ⋅ \hat n$ I have to solve the Laplacian eqn in cylindrical coordinates. Or is there any other way to show it?

Yes, I left out 'l' while typing.

8. Aug 11, 2017

### TSny

For a uniformly polarized dielectric object, the electric field inside and outside the object is completely determined by the surface charge density $\sigma = \vec P \cdot \hat n$, where $\hat n$ is the unit normal at the surface. Thus, if you start with two superimposed objects of equal and opposite uniform volume charge density $\rho$, and you displace them by an infinitesimal amount $d$ in such a way as to produce the same $\sigma$ distribution, then E inside and outside this system will be the same as for the polarized object.

On the left is the dielectric object of arbitrary shape that is uniformly polarized in the horizontal direction. At the small yellow patch of area, there will be a surface charge density $\sigma = \vec P \cdot \hat n = P \cos \theta$.

On the right is the system of displaced objects. The displaced area produces a little volume that you can show has total charge $dq = \rho \, d \cos \theta \, dA$. So, the effective surface charge density is $\sigma = dq/dA = \rho ~ d \cos \theta$.

Thus, the surface charge densities for the two systems will be equal everywhere if you choose $\rho ~ d = P$.