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In summary: 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 $$\mathrm{p}(\mathrm{r})=\sum_\alpha e_\alpha (\mathrm{q}_\alpha -\mathrm{R}) \int_0^\infty \delta(\mathrm{r}-\mathrm{R}-\lambda(\mathrm{q}_\alpha-\mathrm{R}) )d\lambda $$,

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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.

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$$\rho_{\text{pol}}=-\vec{\nabla} \cdot \vec{P}.$$

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.

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$$\vec{p}=\int_V \mathrm{d}^3 r \vec{P}(\vec{r})=V \vec{P}.$$

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https://web.mit.edu/6.013_book/www/chapter6/6.1.html

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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.vanhees71 said:

https://web.mit.edu/6.013_book/www/chapter6/6.1.html

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Thanks!Delta2 said:

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.

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