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?
 A: 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.
A: 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.
