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In Griffith's Electrodynamics, $ \vec P$ is called Polarization (dipole moment per unit volume). Also, the bound volume charges developed in a material in an external field is $\rho_v = - \nabla• \vec P$ and surface charges developed is $\sigma_s = \vec P• \hat n$.

Also, applying gauss's law, we get electric displacement $\vec D$ ($= \epsilon_0 \vec E + \vec P$):

$\nabla • \vec D$ = $\rho_f$

Now, If we have a short cylinder of radius a and length L which carries a uniform polarization P parallel to its Axis then, assuming a cylindrical guassian surface inside the cylinder, as polarization is uniform then there will be no bound charges. Also, As there are no free charges then displacement current is zero everywhere and hence electric field should be: $\vec E = (-1/\epsilon_0) \vec P$ and since if we take a cylindrical Gaussian surface outside the cylinder, then as polarization is zero and displacement current is zero, electric field outside should also be zero. But the book says this is wrong and there will be external $\vec E$ present. How?

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As there are no free charges then displacement current is zero everywhere

This is invalid argument. If free charge density vanishes, then it is true that $$ \nabla \cdot \vec{D} = 0 $$ but this does not imply that $\vec{D}$ vanishes.

Polarized body has electric field, in electrostatic scenario it can be calculated as the Coulomb field of the bound charge density associated with the body. We can calculate this field anywhere as integral over region $V+$ containing the whole body (including its surface charge):

$$ \vec{E}(\mathbf x) = \frac{1}{4\pi \epsilon_0}\int_{V+} \frac{\rho_{bound} (\mathbf x - \mathbf x')}{|\mathbf x - \mathbf x'|^3}d^3\mathbf x'. $$ In terms of polarization $$ \vec{E}(\mathbf x) = \frac{1}{4\pi \epsilon_0}\int_{V+} \frac{-\nabla \cdot \vec{P} (\mathbf x - \mathbf x')}{|\mathbf x - \mathbf x'|^3}d^3\mathbf x'. $$ In your example, there are bound charges on the flat faces of the cylinder. If the faces are at $z=0,z=L$, the bound charge density is $$ \rho_{bound} = -P \delta(z) + P \delta(z-L) $$ where $P$ is magnitude of the polarization.

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