The discussion focuses on the arrangement of polarization in a spherically symmetric manner, specifically addressing the conditions for achieving a uniformly charged sphere with radius R. It establishes that solving the divergence equation in spherical coordinates yields a constant charge density ρp, where the polarization vector P(r) is defined as P_r(r) = - (ρp/3)r. The surface polarization charge density σp is calculated to be σp = - (ρp/3)R, which effectively neutralizes the interior charge density ρp, ensuring the system remains electrically neutral.
PREREQUISITESPhysicists, electrical engineers, and students studying electromagnetism who are interested in the principles of electric polarization and charge distribution in spherical geometries.
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 said: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?
very thanks!Charles Link said: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 ##.