Is the superposition principle a postulate in electrostatics?

Question

Consider two electrical point charges $q_1$ and $q_2$ described by the total charge distribution $\rho = \rho_1 + \rho_2 = q_1 \delta(\vec{r} - \vec{r_1}) + q_2 \delta(\vec{r} - \vec{r_2})$. The total electrical potential could then be calculated by $$ \phi(\vec{r}) = \frac{1}{4 \pi \varepsilon_0} \int_V{\frac{\rho(\vec{r'})}{|\vec{r} - \vec{r'}|} d^3r'} = \frac{1}{4 \pi \varepsilon_0} \int_V{\frac{\rho_1(\vec{r'})}{|\vec{r} - \vec{r'}|} d^3r'} + \frac{1}{4 \pi \varepsilon_0} \int_V{\frac{\rho_2(\vec{r'})}{|\vec{r} - \vec{r'}|} d^3r'} = \phi_1 + \phi_2 $$ as solution of the Poisson's equation, which can be derived from Maxwell's equations.

Since $\vec{E} = - \nabla \phi$ and the gradient operator is linear we have $$\vec{E} = - \nabla \phi = - \nabla (\phi_1 + \phi_2) = - \nabla \phi_1 - \nabla \phi_2 = \vec{E_1} + \vec{E_2}$$ what the superposition principle is claiming. The superposition principle is sometimes called a postulate in electrostatics. But couldn't it be "derived" in that way?

Answer 1

But couldn't it be "derived" in that way?

No, because your starting statement $\rho = \rho_1 + \rho_2$, assumes the superposition principle. Thus using your derivation to justify the superposition principle would be circular reasoning.

Principles, by their very definition, are not derived. Instead we empirically evaluate them with experiments. Theories are then built on these principles. There is no theoretical reason charge densities need add linearly (at least, not that I'm aware of), perhaps in a different universe they don't.

Answer 2

The Maxwell equations contain electrostatics as special case and are linear, so electrostatic's linearity in the electric field is already contained in the Maxwell equations. I see the linearity in the electric field as consequence of $\vec F = q \cdot \vec E$ (where $\vec F$ is a force, $q$ the charge and $\vec E$ the electric field) and the fact that forces (force vectors) are simply added up to give the total force.

Answer 3

Maxwell equations for electrostatics are $\nabla\cdot\vec D = \rho, \nabla\times \vec E = 0$. The second one tells you, that there is a potential field $\phi$ generating electric field as $\vec E = -\nabla \phi$.

Both equations are linear, it means if there is $\rho_1, \rho_2$ satisfying $\nabla\cdot \vec D_1 = \rho_1$ and $\nabla\cdot \vec D_2 = \rho_2$, then $\rho = \rho_1 + \rho_2$ must hold $\nabla\cdot (\vec D_1 + \vec D_2) = \rho$. Equation with circulation is trivial.

The important fact here is, that theory is constructed on Maxwell equations, not on their consequence $\vec E = - \nabla \phi$. The superposition is then another consequency of linearity of differential operators in equations (divergence and circulation). Realize at this point that superposition would hold even if you didn't construct potentials.