Maxwell’s First Equation in Electrostatics

From the Divergence theorem, we have

\overrightarrow{\bigtriangledown }.\overrightarrow{D}= div \overrightarrow{D}=( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})

\lim_{dv->0}\frac{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}{dv}=\lim_{dv->0}( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})

{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}=\( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})dv

from Gauss’s law \oint_{s}\overrightarrow{D}.\overrightarrow{ds} =Q_{enclosed}

{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}=\( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})dv = Q_{enclosed}

dividing it by dv(or) \Delta v differential volume on both sides

 \frac{{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}}{dv} = \frac{Q_{enclosed}}{dv}

by applying limit on both  sides

\lim_{dv->0}\frac{{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}}{dv} = \lim_{dv->0} \frac{Q_{enclosed}}{dv}

div\overrightarrow{D} = \rho _{v}

\overrightarrow{\bigtriangledown }.\overrightarrow{D} = \rho _{v}

This equation is known as Maxwell’s first equation and is also known as point form of Gauss’s law /Differential form of Gauss’s law.



1 Star2 Stars3 Stars4 Stars5 Stars (No Ratings Yet)

Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.