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