Polarization surface charge density on spherical surface - electromagnetics

Problem

A charged metal sphere of radius $$a$$ gives rise to a $$\mathbf{D}$$-field given by $$\mathbf{D}=\mathbf{a_R} \frac{C}{R^2}$$ in the surrounding medium with permittivity $$\varepsilon$$. What is the polarization surface charge density at the spherical surface $$R=a$$?

Attempt

We know the relationship $$\mathbf{P}=\mathbf{D}-\varepsilon_0\mathbf{E}=(\varepsilon-\varepsilon_0)\mathbf{E}$$.

Recalling that $$\mathbf{E}= \frac{\mathbf{D}}{\varepsilon}$$ gives $$\mathbf{P}= \frac{(\varepsilon-\varepsilon_0) \mathbf{D}}{\varepsilon}=\frac{(\varepsilon-\varepsilon_0) C}{\varepsilon a^2}$$

We also have this relationship $$\rho_{\text{ps}}=\mathbf{P} \cdot \mathbf{a_n}$$, where $$\mathbf{a_n}$$ is the outward normal vector.

Questions

It turns out the correct answer to this problem is b), so $$\rho_{\text{ps}}= -\frac{(\varepsilon-\varepsilon_0) C}{\varepsilon a^2}$$ but I don't understand this. The minus-sign must mean that one has to choose a normal vector that points opposite of $$\mathbf{P}$$, but why?

Also, the minus-sign assures that the surface charge density is a negative number, but how can a density be negative? I hope someone can clarify this for me.

1. The "outward normal" for a dielectric object is the vector pointing out of the volume of the object. In this case, since you have a dielectric surrounding a metal sphere, the "outward normal" points from the dielectric into the center of the sphere, and so the outward normal is $$-\hat{r}$$.
1. Charge density $$\rho$$ is defined to be charge per volume. If you have a negative charge in some volume, then $$\rho$$ is negative.
$$Q_p = -\oint \vec{P}\cdot d\vec{A},$$ where $$Q_p$$ is the enclosed net polarisation charge and the RHS is minus the closed surface integral of the polarisation field.