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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 $?

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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.

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2 Answers 2

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  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.
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  • $\begingroup$ Ahh okay. Finally I understand it. Thank you for your help :( $\endgroup$
    – Carl
    Dec 12, 2020 at 20:17
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The electric field set up by the polarisation charges would usually be in the opposite direction to the applied electric field. However, the polarisation field is defined to be in the same direction as the electric field (in linear media). This is the origin of the minus sign.

$$Q_p = -\oint \vec{P}\cdot d\vec{A},$$ where $Q_p$ is the enclosed net polarisation charge and the RHS is minus the closed surface integral of the polarisation field.

Thus if you have an electric field with lines going into a dielectric medium, that will induce negative polarisation charges at the surface even though the polarisation field is in the same direction as the E-field in the medium. Thus the surface charge density is negative.

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