The electric field formed by a uniformly polarized cylinder

[SadStudent]

Homework Statement

We have some uniformly polarized cylinder with some polarization $P\hat{z}$.

The The cylinder's center is on $\hat{z}$ and it's on the $xy$ surface.

The cylinder has radius $R$ and height $d$where $R \gg d$ .

If we were to ground the $xy$ plane what would the electric field be inside the cylinder and outside (disregarding edge effects)?

Homework Equations

Well if the there was no grounded $xy$ surface the electric field would be $-\frac{\overrightarrow{P}}{\varepsilon_{0}}$ inside and outside it would look like a dipole, but now there's is, so I'm not quite sure.

The Attempt at a Solution

I've tried thinking about it in two ways but I couldn't find proof to either of them

• Charges from $\infty$ arrive to "cancel out" the negative surface charge that's on the bottom of the cylinder so it's like a single surface but with field lines only half the magnitude, therefore it's $-\frac{\overrightarrow{P}}{2\varepsilon_{0}}$
  
• Or we can try and find an "image charge" or an "image object" for example if we put an identical cylinder under this cylinder, the upper surface charge on the "image" cylinder will cancel the bottom surface charges of the real one but then we have $-\frac{2\overrightarrow{P}}{\varepsilon_{0}}$ because the $\overrightarrow{P}$ is proportional to the distance.
  

But this is all hand-wavy and there might be a simple way to show which one preserves zero potential on the $xy$ surface but I don't know how to show it mathematically.

 

[TSny]

Hello, SadStudent. Welcome to Physics Forums!

I think your image charge approach is a good way to go. When you say the polarization is uniform and given by $P\vec{z}$, does that mean that $\vec{z}$ is a unit vector?

Anyway, you should be able to easily show that with the image cylinder in place, the potential will be zero everywhere on the xy plane. Pick an arbitrary point on the xy plane. Then pick an arbitrary element of charge on the original cylinder and note that there will be a symmetrically placed element of opposite charge on the image cylinder.

 

[SadStudent]

Thanks, and yes $\vec{z}$ was supposed to be $\hat{z}$ a unit vector I've edited my post accordingly :).
I've figured out the zero potential part, but what about the magnitude of the field would it be doubled, halved or something entirely else ?

I'm confused because :

• We did prove that the magnetic field between two surfaces with $\pm Q$ charge is not proportional to the distance and is $- \frac{\sigma}{\varepsilon_{0}}$, so I could say that this situation is the same with the slight change that $\overrightarrow{P}$ is the surface charge density.
• With an ordinary dipole we've said that $\vec{p} = qd$ but what we did just now with the image cylinder did not change this $d$? or did it ?

I'm still not entirely sure as you can see :) and I'd appreciate a little more help.

 

[TSny]

I've figured out the zero potential part, but what about the magnitude of the field would it be doubled, halved or something entirely else ?

I'm confused because :

[*] We did prove that the magnetic field between two surfaces with $\pm Q$ charge is not proportional to the distance and is $- \frac{\sigma}{\varepsilon_{0}}$, so I could say that this situation is the same with the slight change that $\overrightarrow{P}$ is the surface charge density.

That's right. I'm not quite clear on how far to take the approximations R>>d and "neglect edge effects" . But I think it means that you can treat the upper and lower surfaces of the cylinder as infinite plates of surface charge σ =± P/ε_o for the purpose of finding the electric field inside the material. So, you'll need to think about what happens to the field inside when you go from the case of the original cylinder alone to the case of the original cylinder along with the image cylinder. You are right that the field between two infinite oppositely charged plates does not depend on the position chosen between the plates or the separation distance between the plates. As far as dealing with the field outside, this is where I'm not confident about the approximations. Certainly edge effects would be important if you are considering the field just outside the curved surface of the cylinder.

[*] With an ordinary dipole we've said that $\vec{p} = qd$ but what we did just now with the image cylinder did not change this $d$? or did it ?

The polarization P is given to be uniform in the cylinder. P does not depend on position within the cylinder. An ordinary dipole p is not the same as polarization P. So, although p of an ordinary dipole depends on the distance d between the charges in the dipole, the polarization P is uniform throughout the cylinder. When you are considering the electric field inside the cylinder, you should think of the field as due to the charges ±σ on the large flat surfaces of the cylinder (and the image cylinder) rather than thinking of a simple dipole p = qd.

Now, if you need to consider the electric field at large distance r from the cylinder (r >> R), then you can treat the cylinder as an ordinary dipole p = Qd where d is the height of the cylinder. So, this can help you decide what happens to the distant electric field when you go from the original cylinder alone to the the cylinder sitting on the grounded plane.