# Electric field of infinite cylinder with radial polarization

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?

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.