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Given the electric field defined by: $$\vec {E(\vec r)}= x \hat {x} $$ There is no obvious contradiction with the classic electro-magnetic theory. But: $$\vec \nabla \cdot \vec E = 4 \pi \rho(\vec r) = 1 \Rightarrow \rho=\dfrac {1}{4\pi}$$ Now, this charge density is constant throughout the space, and obviously isotropic. Yet, the field created has certain direction, and it varies in magntitude throughout the space, both last properties are in contradiction to the fact that the charge density is isotropic. Also, other fields, such as: $$\vec E_y = y \hat y $$ will lead to the same charge density function.

I do have some intuition to what is the problem: I think the irrelevancy of such fields and charge densities is the root of the problem. Although they satisfy some fundemental equations of classical physics - they cannot exist in our universe.

Is there another explanation to this?

SchrodingersCat
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Ranc
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1 Answers1

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It is true that given an electric field, then you can define uniquely the charge density that created it, by Gauss' law, as you have done. But the converse is not true: given a charge density you cannot define uniquely the electric field that it will create since you have to solve a differential equation (again Gauss' law) to do that and each differential equation gives a unique solution only if it is accompanied by some boundary conditions. So, the same charge density can produce different electric fields for different boundary conditions.

user3257624
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