Module 2 : Electrostatics
Lecture 11 : Conductors and Dielectric
  The first integral can be converted to a surface integral using the divergence theorem giving,
\begin{displaymath}\phi =\frac{1}{4\pi\epsilon_o} \int_{surface}\frac{\vec P}{ r... ...pi\epsilon_0}\int_{vol} \frac{1}{r}\vec\nabla\cdot\vec P d\tau \end{displaymath}
  The first term is the potential that one would expect for a surface charge density $\sigma_b$where
 
\begin{displaymath}\sigma_b = \vec P\cdot\hat n\end{displaymath}
  where $\hat n$ is the unit vector along outward normal to the surface. The second term is the potential due to a volume charge density $\rho_b$ given by
 
\begin{displaymath}\rho_b = -\vec\nabla\cdot\vec P\end{displaymath}
  The potential due to the dielectric is, therefore, given by
 
\begin{displaymath}\phi =\frac{1}{4\pi\epsilon_o} \int_{surface}\frac{\sigma_b dS}{ r} +\frac{1}{4\pi\epsilon_0}\int_{vol} \frac{\rho_b d\tau}{r} \end{displaymath}
  and the electric field
  \begin{eqnarray*} \vec E &=& -\nabla\phi\\ &=&\frac{1}{4\pi\epsilon_0} \int_{su... ...rac{1}{4\pi\epsilon_0}\int_{vol} \frac{\rho_b \hat r}{r^2}d\tau \end{eqnarray*}
   
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