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[NOTES/EM-03004]-The Electric Stress Tensor

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An expression for the electric stress tensor is derived for a charge distribution in a volume \(V\). The surface integral of the stress  tensor gives the total electric force on the  charge in the volume \(V\).

Given a charge density \(\rho(\vec{r})\), the force in electric field on a small volume \(\Delta V\) is \(\rho \Delta \vec{E}\). If we consider a volume element \(d^3r\) located at \(vec{r}\) is \(\vec{E} \rho d^3r\). The volume integral over \(V\) of this expression will give the total force acting on the volume \(V\) due to the electric field \(\vec{E}\). \begin{equation} \vec{F} = \iiint_V \rho(\vec{r}) \vec{E}(\vec{r}) d^3r\label{EQ01} \end{equation} Using 1,2,3 notation for the components of vectors, we write the \(i^\text{th}\) component of the force as \begin{equation} F_i = \iiint_V \rho(\vec{r}) E_i(\vec{r}) d^3r\label{EQ02} \end{equation} Using Maxwell's equation we can rewrite \(\rho\) in terms of electric field \(\rho=\epsilon_0 \nabla\cdot\vec{E}= \epsilon_0 \sum_j\partial_jE_j\), we get \begin{equation} F_i = \epsilon_0 \iiint \sum_j(\partial_jE_j) E_i\label{EQ03} \end{equation} The Maxwell's equation \(\text{curl} \vec{E}=0\) implies \(\partial_jE_i= \partial_i E_j\) . Therefore \begin{eqnarray} (\partial_jE_j) E_i &=& \partial_j(E_jE_i) - E_j(\partial_j E_i) \\ &=&\partial_j(E_jE_i) - E_j(\partial_i E_j) \label{AGBX03003}\\ &=& \partial_j(E_jE_i) -\frac{1}{2}\partial_i (E_jE_j)\label{EQ04} \end{eqnarray} . Therefore Eq(3) becomes
\begin{eqnarray} F_i &=&\epsilon_0\iiint_V \sum_j \partial_j (E_jE_i) - \frac{1}{2}\,\sum_j\delta_{ij}\partial_j\big( \vec{E}\cdot \vec{E}\big)\\ &=&\epsilon_0\iiint_V \partial_j \Big\{E_iE_j - \frac{1}{2} \delta_{ij}(\vec{E}\cdot\vec{E})\Big\} \end{eqnarray} Using divergence theorem of vector calculus we can write the force \(F_i\) as a surface integral \begin{equation} F_i = \iint_S \Big\{(\hat{n}\cdot\vec{E}) E_i - \frac{1}{2} n_i (\vec{E}\cdot\vec{E})\Big\}\,dS \end{equation} We can write the above equation as vector equation \begin{equation} \boxed{\vec{F} = \iint_S \Big\{(\hat{n}\cdot\vec{E}) \vec{E} - \frac{1}{2} \vec{n} (\vec{E}\cdot\vec{E})\Big\}\,dS} \end{equation} This expression is a surface integral and is to be interpreted as giving the force in terms of pressure per unit area on the boundary surface \(S\).

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