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The Gauss law of electrostatics follows from the Coulomb's law for a point charge and superposition principle. The Gauss law along with the Gauss divergence theorem of vector calculus imply Maxwell's first equation \[ \nabla\cdot\bar{E}=\frac{\rho}{\epsilon_0}. \] for electrostatics.
1. Maxwell's second equation from Coulumb's Law
Consider the field $\vec{E}$ at a point $\vec{r}$ created by a point charge located at $\vec{r}_0$. \begin{eqnarray} \vec{E}&=&\frac{q}{4\pi\epsilon_0}\frac{\vec{r}-\vec{r_0}}{\mid\vec{r}-\vec{r}_0 \mid^3} \label{eq6} \end{eqnarray} That the electric field of a point charge obeys Maxwell's equation \begin{eqnarray} \nabla\times\vec{E}&=&0 \label{eq7} \end{eqnarray} can be seen by direct computation Verify. Remembering superposition principle, for several point charges \begin{eqnarray}\nonumber \vec{E}&=&\sum_k\vec{E_k} \label{eq8} \\ \therefore \qquad\qquad \nabla\times\vec{E}&=&\sum_k\nabla\times\vec{E_k}\\ \nabla\times\vec{E}&=&0 \label{eq10} \end{eqnarray} Thus for several charges,\EqRef{eq10} is true because each term in \eqRef{eq8} is zero.\\ Thus we have \begin{eqnarray} \nonumber\text{Coulomb's law}& \implies& \text{Gauss Law} \\ \nonumber&\implies& \nabla\cdot\vec{E}=\rho/\epsilon_0 \\ \nonumber\text{Coulomb's law}&\implies& \nabla\times\vec{E}=0\\ \nonumber\text{Gauss law}&\not\hspace{-5 pt}\Longrightarrow&\text{Coulomb's law} \end{eqnarray} Though the Gauss law does not imply Coulomb's law, for many special problems with symmetries, it can be used to find the electric field. For example a spherically symmetric charge distribution, the electric field can be obtained using Gauss law and symmetry arguments. We can therefore write
| But Gauss law + symmetry arguments \(\implies\) Electric field |
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4727:Diamond Point
