# Why is the divergence of the field zero in Maxwell's equations?

I read in a book called Vector Analysis by Murray R. Spiegel by Schaums Series, and I found that there is somewhere printed that the divergence of the electric field is zero.

Since my teacher told that divergence means something which originates from a point and meet another point, simply source and sink. And I know that electric field originates from a point charge and in a dipole its sink is the negative charge, then why the divergence of the field is said to be zero in the Maxwell's equations?

When $${\bf \nabla} \cdot {\bf E}$$ is introduced in Vector Analysis by Murray R. Spiegel, it is stated explicitly that it is proportional to the charge density and therefore it is zero only if the charge density is zero.

I guess that you may have been mislead by the solved problem n. 19 of Chapter 4, where it is shown that $${\bf \nabla} \cdot \left( \frac{{\bf r}}{r^3}\right)=0.$$ I.e. that the divergence of a Coulomb-like field would be zero.

In that case, you have to be careful. The equality holds at the points where the function is differentiable. i.e. everywhere but the origin ($$\bf r = 0$$). At the origin, that vector function is singular and its divergence can be evaluated, within distribution theory, only as a generalized Dirac delta function $$\delta({\bf r})$$. For more details, have a look at this Q&A on Math.SE.

• So it would be more correct to write ${\bf \nabla} \cdot \left( \frac{{\bf r}}{r^3}\right)=4 \pi \delta(\bf r)$. – Thomas Fritsch Dec 2 at 10:51
• @ThomasFritsch Sure, if one allows generalized functions. But confining the result to ordinary function the value of the divergence is zero at all points excluded the origin. – GiorgioP Dec 2 at 11:07
• @GiorgioP Please can you explain it in detail that why the divergence is zero at all points excluding the origin? – Diku Khanikar Dec 2 at 11:17
• @DikuKhanikar at each point ${\bf r}\neq {\bf 0}$, divergence is zero because $\nabla \cdot \left( \frac{{\bf r}}{r^3} \right) = (\nabla \cdot {\bf r} )\frac{1}{r^3}+{\bf r} \cdot \nabla \left( \frac{{1}}{r^3} \right)= \frac{3}{r^3}-3 \frac{{\bf r}\cdot {\bf r}}{r^5}= 0$. At ${\bf r}={\bf 0}$ the Coulomb field is not defined. – GiorgioP Dec 2 at 13:08
• @DikuKhanikar From the other side of Gauss's Law, divergence is zero everywhere except the origin because the Coulomb field is generated by a single point charge at the origin. This means there is no charge anywhere except the origin, so $\rho=0$ everywhere except the origin. At the origin, $\rho$ is infinite, but it's infinite in a particular way such that its spatial integral is finite (and is exactly the magnitude of the point charge). – probably_someone Dec 2 at 16:09

It's not. Gauss's Law (which is one of Maxwell's Equations) states that

$$\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}$$

As for why the book allegedly said it's zero, it might have been referring to the electric field in a vacuum, where $$\rho=0$$ by definition. Without a more specific reference than "somewhere in the book", it's not possible to verify this, though.

• $\rho$ is the charge density of the medium. – FakeMod Dec 2 at 10:01