Consider a small closed box, with sides parallel to the coordinate planes. What is the flux of $\EE$ out of the box?

Consider first the vertical contribution, namely the flux up through the top plus the flux down through the bottom. These two sides each have area element $dA=dx\,dy$, but the outward normal vectors point in opposite directions, so we get \begin{eqnarray*} \sum_{\rm top+bottom} \EE \cdot d\AA &=& \EE(z+dz) \cdot \kk\> dx\,dy - \EE(z) \cdot\kk\> dx\,dy \\ &=& \Bigl( E_z(z+dz) - E_z(z) \Bigr) \> dx\,dy \\ &=& \frac{E_z(z+dz) - E_z(z)}{dz} \> dx\,dy\,dz \\ &=& \Partial{E_z}{z} \>\> dx\,dy\,dz \end{eqnarray*}

Repeating this argument using the remaining pairs of faces, it follows that the total flux out of the box is \begin{eqnarray*} {\rm total flux} = \sum_{\rm box} \EE \cdot d\AA = \left( \Partial{E_x}{x} + \Partial{E_y}{y} + \Partial{E_z}{z} \right) \> d\tau \end{eqnarray*} Since this is proportional to the volume of the box, it approaches zero as the box shrinks down to a point. The interesting quantity is therefore the ratio of the flux to volume. This is the divergence.

At any point $P$, we therefore define the divergence of a vector field $\EE$, written $\grad\cdot\EE$, to be the flux of $\EE$ per unit volume leaving a small box around $P$. In other words, the divergence is the limit as the box collapses around $P$ of the ratio of the flux of the electric field out of the box to the volume of the box. Thus, the divergence of $\EE$ at $P$ is the flux per unit volume through a small box around $P$, which is given in rectangular coordinates by \begin{eqnarray*} \grad\cdot\EE = \frac{\rm flux}{\rm unit volume} = \Partial{E_x}{x} + \Partial{E_y}{y} + \Partial{E_z}{z} \end{eqnarray*}

You may have seen this formula before, but remember that it is merely the rectangular coordinate expression for the divergence of $\EE$; the divergence is defined as flux per unit volume. Similar computations can be used to determine expressions for the divergence in other coordinate systems.