Coulomb's Law from vector identities

[ralqs]

I can show that Coulomb's Law + superposition implies $\nabla \cdot \mathcal {E} = \frac{\rho}{\epsilon_0}$ and $\nabla \times \mathcal{E} = \mathbf{0}$. I want to go the other way and derive Coulomb's law and superposition from the vector identities. I know that Gauss' Law implies Coulomb's law if we assume that the electric field is radial and symmetric. Can these assumptions be justified from $\nabla \times \mathcal{E} = \mathbf{0}$? Thanks.

 

[henry_m]

We can expect the field to be radially symmetric on purely physical grounds, in which case curl E=0 seems a bit redundant.

But from a mathematical point of view, without curl E=0 there's a massive problem of nonuniqueness to the solutions: we can add an extra arbitrary field as long as it has zero divergence, so there are loads of possible solutions (though only one spherically symmetric).

Including curl E=0 mostly solves this problem: it means you can write E=-grad phi, where phi is the electrostatic potential. But there is still a nonuniqueness issue, because you can add to phi any solution of Laplace's equation. For example, we could have an extra constant magnetic field in the x-direction.

There's one more ingredient needed: the boundary conditions. If we specify that we want the field to tend to zero as we go to infinity, we get uniqueness. Only then can we guarantee that the field will be radially symmetric.

 

[ralqs]

There's one more ingredient needed: the boundary conditions. If we specify that we want the field to tend to zero as we go to infinity, we get uniqueness. Only then can we guarantee that the field will be radially symmetric.

Are you sure? Doesn't Gauss' Law guarantee that the force tends to zero as we tend to infinity?

But anyways, I managed to prove what I wanted. Thanks!

 

[henry_m]

Are you sure? Doesn't Gauss' Law guarantee that the force tends to zero as we tend to infinity?

Unfortunately not: for example, a constant electric field will have zero flux through any closed surface, but it certainly doesn't disappear at infinity. And there are infinitely many more nontrivial examples: pick any solution to Laplace's equation and take its gradient.

These can all be ruled out usually on physical grounds (unless you're interested, for example, in a situation where there is an externally applied field), which mathematically correspond to boundary conditions.

 

[sardar]

Thank you for your question. It is certainly possible to derive Coulomb's Law and superposition from the vector identities \nabla \cdot \mathcal {E} = \frac{\rho}{\epsilon_0} and \nabla \times \mathcal{E} = \mathbf{0}. In fact, these identities are fundamental in understanding the behavior of electric fields and are commonly used in electromagnetism.

To answer your question, the assumptions of a radial and symmetric electric field can indeed be justified from \nabla \times \mathcal{E} = \mathbf{0}. This is because if the electric field is radially symmetric, it means that its magnitude only depends on the distance from the source and not on the direction. This leads to \nabla \times \mathcal{E} = \mathbf{0} since the curl of a radially symmetric vector field is always zero.

Similarly, if we assume that the electric field is radial, it means that it only has a component in the direction towards or away from the source. This leads to \nabla \cdot \mathcal{E} = \frac{\rho}{\epsilon_0} since the divergence of a radially symmetric vector field is proportional to the source charge density.

Using these assumptions, we can then derive Coulomb's Law and superposition from the vector identities. This is because Coulomb's Law states that the electric field at a point is proportional to the source charge and inversely proportional to the distance from the source, which is consistent with the assumptions we have made. Superposition can also be derived from the vector identities by considering the electric field due to multiple point charges.

In conclusion, the vector identities \nabla \cdot \mathcal {E} = \frac{\rho}{\epsilon_0} and \nabla \times \mathcal{E} = \mathbf{0} are powerful tools that can be used to derive Coulomb's Law and superposition. The assumptions of a radial and symmetric electric field can be justified from these identities, making them an essential part of understanding the behavior of electric fields.