Deriving Gauss's law from Coulomb's law

Deriving Gauss's law from Coulomb's law

Gauss's law can be derived from Coulomb's law, which states that the electric field due to a stationary point charge is:

\mathbf{E}(\mathbf{r}) = \frac{q}{4\pi \epsilon_0} \frac{\mathbf{e_r}}{r^2}


er is the radial unit vector,
r is the radius, |r|,
ε0 is the electric constant,
q is the charge of the particle, which is assumed to be located at the origin.

Using the expression from Coulomb's law, we get the total field at r by using an integral to sum the field at r due to the infinitesimal charge at each other point s in space, to give

\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(\mathbf{s})(\mathbf{r}-\mathbf{s})}{|\mathbf{r}-\mathbf{s}|^3} d^3 \mathbf{s}

where ρ is the charge density. If we take the divergence of both sides of this equation with respect to r, and use the known theorem

\nabla \cdot \left(\frac{\mathbf{s}}{|\mathbf{s}|^3}\right) = 4\pi \delta(\mathbf{s})

where δ(s) is the Dirac delta function, the result is

\nabla\cdot\mathbf{E}(\mathbf{r}) = \frac{1}{\epsilon_0} \int \rho(\mathbf{s})\ \delta(\mathbf{r}-\mathbf{s})\ d^3 \mathbf{s}

Using the "sifting property" of the Dirac delta function, we arrive at

\nabla\cdot\mathbf{E}(\mathbf{r}) = \rho(\mathbf{r})/\epsilon_0

which is the differential form of Gauss's law, as desired.

Note that since Coulomb's law only applies to stationary charges, there is no reason to expect Gauss's law to hold for moving charges based on this derivation alone. In fact, Gauss's law does hold for moving charges, and in this respect Gauss's law is more general than Coulomb's law.

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Mon, 09/14/2009 - 11:55