The bound surface charge in a dielectric is given as $\sigma_{b}=\vec{P} \cdot \hat{n}$. Where $\vec{P}$ is the polarisation and $\hat{n}$ is the surface normal. Could anyone please give me a intuitive explanation for this?.


1 Answer 1


The following analogy may help intuition, although that's very subjective.

  1. Remember that a discontinuity in the electric field at a surface is a sign of the presence of a surface charge density, which is given by $$\sigma=\varepsilon_0[{\bf E}_{\rm above}-{\bf E}_{\rm below}]\cdot\hat{\bf n},$$ where $\hat{\bf n}$ is the normal vector to the surface (pointing from below to above). This was a consequence of the electric field having divergence $\nabla\cdot\bf E=\rho/\varepsilon_0$.

  2. In an analogous way, the polarization vector $\bf P$ has divergence $\nabla\cdot{\bf P}=-\rho_b$, or written in a more suggestive way, $\nabla\cdot{\bf P}=\rho_b/(-1)$; therefore a discontinuity in the polarization vector at a surface causes a bound surface charge given by $$\sigma_b=-[{\bf P}_{\rm above}-{\bf P}_{\rm below}]\cdot\hat{\bf n}.$$ If we take the surface to be the outer surface of a dielectric and the normal $\hat{\bf n}$ pointing from the inside of the dielectric to the outside, then ${\bf P}_{\rm above}=0$ (because outside we don't have dielectric) and ${\bf P}_{\rm below}$ is the $\bf P$ of the dielectric, hence $$\sigma_b=-[-{\bf P}_{\rm below}]\cdot\hat{\bf n}={\bf P}\cdot\hat{\bf n}.$$


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