text: # Homework Help: Bound volume and surface charges in dielectric
  
  1. May 6, 2014
  
  ### unscientific
  
  1. The problem statement, all variables and given/known data
  
  Find surface and volume charge densities. Deduce electric field.
  
  2. Relevant equations
  
  3. 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:
  $$\epsilon_0 E (4\pi r^2) = \int_0^a \rho_f dr$$
  $$E = \frac{\epsilon_r P_0}{4\pi \epsilon_0 (\epsilon_r -1)} \left(\frac{a}{r}\right)^2$$
  
  Last edited: May 6, 2014
  2. May 7, 2014
  
  ### 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.
  
  3. May 26, 2014
  
  ### unscientific
  
  Ok, so the bound volume chage density is $-\nabla .\vec{P}$.
  
  The surface charge density is $\vec P \cdot \hat n$.
  
  $$\epsilon_0 \nabla \cdot \vec E = \rho_b + \rho_f$$
  $$\epsilon_0 \nabla \cdot \vec E = \rho_b = -\nabla \cdot \vec P$$
  Thus, $E = \frac{1}{\epsilon_0} \vec P$ ?
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