# Intuitive explanation for the formula of bound surface charge in a dielectric

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. 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 $$\sigma_b=-[-{\bf P}_{\rm below}]\cdot\hat{\bf n}={\bf P}\cdot\hat{\bf n}.$$