Bound volume and surface charges in dielectric

[unscientific]

Homework Statement

Find surface and volume charge densities. Deduce electric field.

Homework Equations

The Attempt at a Solution

Volume charge density:
$$\epsilon_0 \epsilon_r \nabla . \vec E = \rho_f$$

Using $\vec P = \chi \epsilon_0 \vec E = (\epsilon_r -1)\epsilon_0 \vec E$:
$$\left(\frac{\epsilon_r}{\epsilon_r -1}\right) \nabla . \vec P = \rho_f$$

Thus volume charge density:
$$\rho_f = \left(\frac{\epsilon_r}{\epsilon_r -1}\right) \frac{1}{r^2}\frac{\partial}{\partial r} \left[ P_0 r^3(a-r)\right]$$
$$\rho_f = \left(\frac{\epsilon_r}{\epsilon_r -1}\right) P_0 (3a-4r)$$

Surface charge density is less tedious:
$$\sigma_b = \vec P . \hat n = P_0 r(a-r)$$

Isn't the electric field within the sphere simply $\vec E = \frac{1}{\epsilon_0 (\epsilon_r -1)} \vec P = \frac{P_0}{\epsilon_0 (\epsilon_r -1)} r(a-r) \hat r$?

For electric field outside sphere:
$$\epsilon_0 E (4\pi r^2) = \int_0^a \rho_f dr$$
$$E = \frac{\epsilon_r P_0}{4\pi \epsilon_0 (\epsilon_r -1)} \left(\frac{a}{r}\right)^2$$

 

[Sunil Simha]

Hi,

Isn't volume charge density simply $-\nabla .\vec{P}$? And since they have not mentioned $\epsilon_r$, you cannot use it in your solution. Moreover, they have not said that the material in linear, isotropic and homogeneous... so $\vec{P}$ and $\vec{E}$ may not be related as such.

 

[unscientific]

Hi,

Isn't volume charge density simply $-\nabla .\vec{P}$? And since they have not mentioned $\epsilon_r$, you cannot use it in your solution. Moreover, they have not said that the material in linear, isotropic and homogeneous... so $\vec{P}$ and $\vec{E}$ may not be related as such.

Ok, so the bound volume chage density is $-\nabla .\vec{P}$.

The surface charge density is $\vec P \cdot \hat n$.

Gauss's law reads:

$$\epsilon_0 \nabla \cdot \vec E = \rho_b + \rho_f$$

In this case there are no free charges, so:

$$\epsilon_0 \nabla \cdot \vec E = \rho_b = -\nabla \cdot \vec P$$

Thus, $E = \frac{1}{\epsilon_0} \vec P$ ?