Suppose an electric field E(x, y, z) has the form
E_{x}=a x, \quad E_{y}=0, \quad E_{z}=0where a is a constant. What is the charge density? How do you account for the fact that the field points in a particular direction, when the charge density is uniform? [This is a more subtle problem than it looks, and worthy of careful thought.]
\rho=\epsilon_{0} \nabla \cdot E =\epsilon_{0} \frac{\partial}{\partial x}(a x)=\epsilon_{0} a (constant everywhere).
The same charge density would be compatible (as far as Gauss’s law is concerned) with E =a y \hat{ y }, for instance, orE =\left(\frac{a}{3}\right) r, etc. The point is that Gauss’s law (and ∇×E = 0) by themselves do not determine the field—like any di↵erential equations, they must be supplemented by appropriate boundary conditions. Ordinarily, these are so “obvious” that we impose them almost subconsciously (“E must go to zero far from the source charges”)—or we appeal to symmetry to resolve the ambiguity (“the field must be the same—in magnitude—on both sides of an infinite plane of surface charge”). But in this case there are no natural boundary conditions, and no persuasive symmetry conditions, to fix the answer. The question “What is the electric field produced by a uniform charge density filling all of space?” is simply ill-posed: it does not give us sufficient information to determine the answer. (Incidentally, it won’t help to appeal to Coulomb’s law \left( E =\frac{1}{4 \pi \epsilon_{0}} \int \rho \frac{\hat{ᴫ} }{ᴫ^{2}} d \tau\right)—the integral is hopelessly indefinite, in this case.)