Spherically arranged polarization

[Sang-Hyeon Han]

If we arrange the polarization spherically, Does it make a uniformly charged sphere radius of R?? If right, How can I find out \vec{P}(\vec{r}) which results in the constant charge density \rho?

 

[Charles Link]

If we arrange the polarization spherically, Does it make a uniformly charged sphere radius of R?? If right, How can I find out \vec{P}(\vec{r}) which results in the constant charge density \rho?

You can solve the divergence equation in spherical coordinates to produce a $\rho_p$ that is constant everywhere, where $-\nabla \cdot P=\rho_p$, but polarization still creates an electrically neutral system, so that the surface polarization charge density of $\sigma_p=P \cdot \hat{n}$ will be such that it neutralizes any positive $\rho_p$ in the interior. Solving the divergence equation, you get $P_r(r)=-(\rho_p/3)r$. $\\$ $\int \rho_p dv=+(4/3) \pi R^3 \rho_p.$ $\\$ Now $P_r=-(\rho_p/3)R$ at $r=R$, so that $\sigma_p=-(\rho_p/3)R$. $\\$ $\int \sigma_p dA=-(\rho_p/3)R(4 \pi R^2)$ precisely neutralizing $\int \rho_p dv$.

 

[Sang-Hyeon Han]

You can solve the divergence equation in spherical coordinates to produce a $\rho_p$ that is constant everywhere, where $-\nabla \cdot P=\rho_p$, but polarization still creates an electrically neutral system, so that the surface polarization charge density of $\sigma_p=P \cdot \hat{n}$ will be such that it neutralizes any positive $\rho_p$ in the interior. Solving the divergence equation, you get $P_r(r)=-(\rho_p/3)r$. $\\$ $\int \rho_p dv=+(4/3) \pi R^3 \rho_p.$ $\\$ Now $P_r=-(\rho_p/3)R$ at $r=R$, so that $\sigma_p=-(\rho_p/3)R$. $\\$ $\int \sigma_p dA=-(\rho_p/3)R(4 \pi R^2)$ precisely neutralizing $\int \rho_p dv$.

very thanks!