Dipole moment of a cylinder of uniform polarization

  • #1
Ahmed1029
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If I want to calculate the dipole moment of a dielectric cylinder of uniform polarization perpendicular to its axis, I could multiply the polarization by the volume of the cylinder, which is okay. But another method is to consider the cylinder to be a superposition of two cylinders of equal and opposite volume charge densities, with the two axes separated by a distance d, and then calculate the dipole moment by multiplying the d vector by the charge of the positive cylinder. But this is the same way we obtain the dipole moment of two point charges. I've been looking for some derivation or justification, but I couldn't find any. Could someone justify this method for me?
 

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  • #2
Delta2
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You can view the polarized cylinder as consisting of many dipoles each with dipole moment ##\vec{p_i}=q_i\vec{d_i}##. Then it is a straightforward result (look https://en.wikipedia.org/wiki/Electric_dipole_moment#Expression_(general_case))

that the total dipole moment is equal to $$\vec{p}=\sum_i \vec{p_i}=\sum_i q_i\vec{d_i}$$

Now if the polarization is uniform then we can set ##\vec{d_i}=\vec{d}=constant## so we shall have
$$\vec{p}=\sum_i q_i \vec{d}=Q\vec{d}$$ where Q the total positive charge of the many dipoles. Of course the cylinder is neutral so its total charge is ##Q-Q=0##, but Q is the charge of the positively charged semi cylinder in what you mention in the OP.
 
  • #3
vanhees71
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Note that a polarization of a body is equivalent to a charge density
$$\rho_{\text{pol}}=-\vec{\nabla} \cdot \vec{P}.$$
For a uniformly polarized body that results in ##\rho_{\text{pol}} \in V \setminus \partial V## and a surface-charge distribution
$$\sigma_{\text{pol}}|_{\partial V}=-\vec{n} \cdot (\vec{P}_1-\vec{P}_2)=\vec{n} \cdot \vec{P}_2,$$
where ##\partial V## is the boundary of the volume, ##\vec{n}## is the surface normal vector pointing out of the volume, and ##\vec{P}_1## is the value of the polarization as places along the surface when taking the limit from outside of the volume (in the present case, ##\vec{P}_1=0##) and ##\vec{P}_2## the value taking the limit from inside (in the present case that's the constant value inside the cylinder).

You can, in principle, calculate the electrostatic potential from this in the usual way,
$$\Phi(\vec{r})=\int_{\partial V} \mathrm{d}^2 \vec{f}' \cdot \vec{P} \frac{1}{4 \pi \epsilon_0 |\vec{r}-\vec{r}'|}.$$
I'm not sure whether this will be possible in closed form though.
 
  • #4
Delta2
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sorry @vanhees71 I think you missed the point here, can you give us hints on how to prove that the dipole moment of a cylinder with uniform polarization ##\vec{P}## is $$\vec{p}=V\vec{P}$$ where V the volume of the cylinder
 
  • #5
vanhees71
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That's just the definition of the dipole-moment density, i.e., ##\vec{P}(\vec{r})## is the "electric dipole moment per unit volume". If ##\vec{P}=\text{const}##, then the total dipole moment is
$$\vec{p}=\int_V \mathrm{d}^3 r \vec{P}(\vec{r})=V \vec{P}.$$
 
  • #6
Delta2
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eeer, sorry are you saying that the polarization can be defined as electric dipole moment per unit volume? This is not the standard definition of polarization, I can tell.
 
  • #8
Delta2
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Yes sorry I thought it was defined from ##\vec{D}## and ##\vec{E}## via the equation $$\vec{D}=\epsilon_0\vec{E}+\vec{P}$$, but that is rather the definition of D o:)
 
  • #9
DrDu
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This IS the standard definition of polarization. What else do you think it is? Do you have a reference, where it's claimed otherwise? Here's a nice treatment:

https://web.mit.edu/6.013_book/www/chapter6/6.1.html
I remember that even for localized charge distributions the polarization will involve higher order multipole moments. In principle, P is fixed by the condition that it's divergence equals the charge distribution up to the rotation of an arbitrary field, as div rot =0. The irrotational part may be obtained as $$\mathrm{p}(\mathrm{r})=\sum_\alpha e_\alpha (\mathrm{q}_\alpha -\mathrm{R}) \int_0^\infty \delta(\mathrm{r}-\mathrm{R}-\lambda(\mathrm{q}_\alpha-\mathrm{R}) )d\lambda $$, where ##\mathrm{q}_\alpha## is the location of the individual charges ##e_\mathrm{\alpha}## and ##\mathrm{R}## the center of the multipole expansion, see D. P. Craig, T. Thirunamachanrdran, Molecular Quantum Electrodynamics, Dover Pubs, Eq. 10.2.10. Clearly, the dipole moment density is only a lowest order approximation. The choice of the rotational part depends on convention. In electrostatics, it is usually assumed to vanish, but in optics, it makes sense to use it to incorporate magnetic effects.
 
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  • #10
Ahmed1029
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You can view the polarized cylinder as consisting of many dipoles each with dipole moment ##\vec{p_i}=q_i\vec{d_i}##. Then it is a straightforward result (look https://en.wikipedia.org/wiki/Electric_dipole_moment#Expression_(general_case))

that the total dipole moment is equal to $$\vec{p}=\sum_i \vec{p_i}=\sum_i q_i\vec{d_i}$$

Now if the polarization is uniform then we can set ##\vec{d_i}=\vec{d}=constant## so we shall have
$$\vec{p}=\sum_i q_i \vec{d}=Q\vec{d}$$ where Q the total positive charge of the many dipoles. Of course the cylinder is neutral so its total charge is ##Q-Q=0##, but Q is the charge of the positively charged semi cylinder in what you mention in the OP.
Thanks!
 
  • #11
DrDu
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I think one confusion with the definition of polarisation is that, while it can in deed be described as a density of hypothetical infinitesimal dipoles, it is in general not sufficient to only consider the real dipole moments of the atoms or molecules making up the substance, although sometimes this may be a good approximation.
 

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