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In Griffiths E&M textbook, there is a problem regarding finding the electric field inside and outside of an infinite cylinder with uniform polarization $\bf{P}$ and radius $a$. I'm thinking about an infinite cylinder with radial polarization, so $\textbf{P} = P_0 s \hat{s}$ where $P_0$ is just some constant polarization factor. So, to do this you find the bound charges. The bound surface charge is $\sigma = \textbf{P} \dot{} \hat{n} = P_0 s \hat{s}\dot{}\hat{s} = P_0 s$. The bound volume charge is $\rho = - \bigtriangledown\dot{}\textbf{P}=-\frac{1}{s}\frac{d}{ds}s P_0 s = -2P_0$. Then Gauss's law can be applied because there is cylindrical symmetry. So to find the electric field inside, you have a gaussian cylinder of radius $r$ and do $\int \textbf{E}\dot{}dA=\frac{Q}{\epsilon_0}$ which can be simplified to be $\textbf{E}(2\pi r L) = -\frac{2P_0}{\epsilon_0}(\pi L r^2)$, then $\textbf{E}_{inside} = -\frac{P_0}{\epsilon_0}r\hat{s}$. The outside is solved in the same way by now $Q = \int_0^a\rho d\tau + 2\pi a L P_0 s$ because it is the total volume bound charge plus the total area of the surface bound charge. And this solves to be $Q = -2 P_0 \pi L a^2 + 2 P_0 \pi L a^2 =0$. So $\textbf{E}_{outside} = 0$. This doesn't make sense though, shouldn't there be a charge outside since the cylinder is charged?

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The cylinder is locally charged, but the charges on the surface and the bulk volume cancel out. You can think of it being qualitatively similar to how if you have a point charge of charge $Q$ surrounded by a spherical shell of total charge $-Q$ uniformly distributed, it will have no electric field outside the shell.

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