Is Gauss's Law in Differential Form Dependent on Position?

[ehrenfest]

Homework Statement

Gauss's Law is often given as:

\nabla \cdot \vec{E} = \rho/ \epsilon_0

However E is, in general a function of position, so the equation is really
\nabla \cdot \vec{E}(\vec{r}) = \rho(\vec{r}) /\epsilon_0
correct?

Homework Equations

The Attempt at a Solution

 

[pam]

Yes. The (r) is often left out, but understood.
Just apply the divrgence theorem to get Gauss's integral law.