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Mathematically speaking, it is true that Maxwell's equations by themselves aren't the whole story; they're a set of PDEs for which one needs to specify boundary conditions separately if one wants to solve them. Boundary conditions can be well-motivated from a physical perspective in a given scenario, but they do not follow from the equations themselves.

As for Poisson's equation versus Coulomb's Law, no requirement of spherical symmetry is necessary. Start with Poisson's equation, and set the charge density to be that of a point charge, namely \begin{align} \rho(\mathbf x) = q\delta(\mathbf x - \mathbf x_0). \end{align} Next, use the vanishing curl of the electric field so write $\mathbf E = -\nabla \Phi$ and plug this into Poisson's equation to obtain \begin{align} \nabla^2\Phi = -\frac{q}{\epsilon_0}\delta(\mathbf x-\mathbf x_0) \end{align} In other words, we want to determine $\Phi$ that is a Green's function for the Laplace equation. The general solution is \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} + F(\mathbf x) \end{align} where $F$ is a harmonic function, namely one which satisfies Laplace's equation. If we then invoke the boundary condition that the potential vanishes at infinity, then this forces $F$ to be identically zero, and we obtain \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} \end{align} which, upon taking the gradient, yields he electric field of a point charge which is essentially the content of Coulomb's Law.

Mathematically speaking, it is true that Maxwell's equations by themselves aren't the whole story; they're a set of PDEs for which one needs to specify boundary conditions separately if one wants to solve them. Boundary conditions can be well-motivated from a physical perspective in a given scenario, but they do not follow from the equations themselves.

As for Poisson's equation versus Coulomb's Law, no requirement of spherical symmetry is necessary. Start with Poisson's equation, and set the charge density to be that of a point charge, namely \begin{align} \rho(\mathbf x) = q\delta(\mathbf x - \mathbf x_0). \end{align} Next, use the vanishing curl of the electric field (which works in the absence of explicitly time-varying magnetic fields) to write $\mathbf E = -\nabla \Phi$ and plug this into Poisson's equation to obtain \begin{align} \nabla^2\Phi = -\frac{q}{\epsilon_0}\delta(\mathbf x-\mathbf x_0) \end{align} In other words, we want to determine $\Phi$ that is a Green's function for the Laplace equation. The general solution is \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} + F(\mathbf x) \end{align} where $F$ is a harmonic function, namely one which satisfies Laplace's equation. If we then invoke the boundary condition that the potential vanishes at infinity, then this forces $F$ to be identically zero, and we obtain \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} \end{align} which, upon taking the gradient, yields he electric field of a point charge which is essentially the content of Coulomb's Law.

Mathematically speaking, it is true that Maxwell's equations by themselves aren't the whole story; they're a set of PDEs for which one needs to specify boundary conditions separately if one wants to solve them. Boundary conditions can be well-motivated from a physical perspective in a given scenario, but they do not follow from the equations themselves.

As for Poisson's equation versus Coulomb's Law, no requirement of spherical symmetry is necessary. Start with Poisson's equation, and set the charge density to be that of a point charge, namely \begin{align} \rho(\mathbf x) = q\delta(\mathbf x - \mathbf x_0). \end{align} Next, use the vanishing curl of the electric field so write $\mathbf E = -\nabla \Phi$ and plug this into Poisson's equation to obtain \begin{align} \nabla^2\Phi = -\frac{q}{\epsilon_0}\delta(\mathbf x-\mathbf x_0) \end{align} In other words, we want to determine $\Phi$ that is a Green's function for the Laplace equation. The general solution is \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} + F(\mathbf x) \end{align} where $F$ is a harmonic function, namely one which satisfies Laplace's equation. If we then invoke the boundary condition that the potential vanishes at infinity sufficiently rapidly, then this forces $F$ to be identically zero, and we obtain \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} \end{align} which, upon taking the gradient, yields he electric field of a point charge which is essentially the content of Coulomb's Law.

Mathematically speaking, it is true that Maxwell's equations by themselves aren't the whole story; they're a set of PDEs for which one needs to specify boundary conditions separately if one wants to solve them. Boundary conditions can be well-motivated from a physical perspective in a given scenario, but they do not follow from the equations themselves.

As for Poisson's equation versus Coulomb's Law, no requirement of spherical symmetry is necessary. Start with Poisson's equation, and set the charge density to be that of a point charge, namely \begin{align} \rho(\mathbf x) = q\delta(\mathbf x - \mathbf x_0). \end{align} Next, use the vanishing curl of the electric field so write $\mathbf E = -\nabla \Phi$ and plug this into Poisson's equation to obtain \begin{align} \nabla^2\Phi = -\frac{q}{\epsilon_0}\delta(\mathbf x-\mathbf x_0) \end{align} In other words, we want to determine $\Phi$ that is a Green's function for the Laplace equation. The general solution is \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} + F(\mathbf x) \end{align} where $F$ is a harmonic function, namely one which satisfies Laplace's equation. If we then invoke the boundary condition that the potential vanishes at infinity, then this forces $F$ to be identically zero, and we obtain \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} \end{align} which, upon taking the gradient, yields he electric field of a point charge which is essentially the content of Coulomb's Law.

Mathematically speaking, it is true that Maxwell's equations by themselves aren't the whole story; they're a set of PDEs for which one needs to specify boundary conditions separately if one wants to solve them. Boundary conditions can be well-motivated from a physical perspective in a given scenario, but they do not follow from the equations themselves.

As for Poisson's equation versus Coulomb's Law, no requirement of spherical symmetry is necessary. If one starts with Poisson's equation and integrates both sides, over some volume $V$, then one simply uses the divergence theorem to translate the integral of the divergence of the electric field into a surface integral over the boundary $\partial V$ of that volume \begin{align} \int_V \nabla\cdot \mathbf E = \int_{\partial V} \mathbf E\cdot d\mathbf a \end{align} and the integral of the charge density over this volume is simply the total charge inside; \begin{align} \int_V \frac{\rho}{\epsilon_0} \,d^3x = \frac{Q_{\mathrm{enc}}}{\epsilon_0} \end{align} and Coulomb's Law in integral form follows immediately.

Mathematically speaking, it is true that Maxwell's equations by themselves aren't the whole story; they're a set of PDEs for which one needs to specify boundary conditions separately if one wants to solve them. Boundary conditions can be well-motivated from a physical perspective in a given scenario, but they do not follow from the equations themselves.

As for Poisson's equation versus Coulomb's Law, no requirement of spherical symmetry is necessary. Start with Poisson's equation, and set the charge density to be that of a point charge, namely \begin{align} \rho(\mathbf x) = q\delta(\mathbf x - \mathbf x_0). \end{align} Next, use the vanishing curl of the electric field so write $\mathbf E = -\nabla \Phi$ and plug this into Poisson's equation to obtain \begin{align} \nabla^2\Phi = -\frac{q}{\epsilon_0}\delta(\mathbf x-\mathbf x_0) \end{align} In other words, we want to determine $\Phi$ that is a Green's function for the Laplace equation. The general solution is \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} + F(\mathbf x) \end{align} where $F$ is a harmonic function, namely one which satisfies Laplace's equation. If we then invoke the boundary condition that the potential vanishes at infinity sufficiently rapidly, then this forces $F$ to be identically zero, and we obtain \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} \end{align} which, upon taking the gradient, yields he electric field of a point charge which is essentially the content of Coulomb's Law.