-3

Everywhere it is written that electrons flow through through wire of an eletric circuit, and the reason given is that the battery maintains high potential at one end, and low at another. But, if you think about it from first principle electricity& magnetism, then one would realize by intuition that the provided reason only would imply that the electron would move from the high potential region to the low potential region, not that it takes the specific route of the wire.

To find a specific route, it would involve solving of the differential equations. In regards to solution, many places I've seen have written that the eletric field caused by the battery only exists inside the wire, but how do we know that/justify this approximation?

I have found this related question, but I felt the answers quite unsatisfactory in regards to the above problem. Also, I Understand that to study circuits, we need more than Maxwell's equations, that is a theory of the matter (charges involved and such) as discussed in this question, but my question is, what is the basis for the justifying that the eletric field is confined to the wire?

Clemens Bartholdy
  • 8,309

1 Answers1

3

The electric field certainly doesn't exist only in the wire. This is easily seen by realizing that $$V_\text{battery} = \int\limits_{+\text{ terminal}}^{-\text{ terminal}}\vec E\cdot d\vec \ell$$ and choosing a path of integration not along the wire.

Current flows in wires and not through the air because air isn't conductive. Ohm's law is usually a pretty good description of how current density relates to the E-field in metals: $$\vec J = \sigma \vec E$$ where $\sigma$ is the electical conductivity. This also works for air except at very high fields, with $\sigma=0$. Metals in contrast generally have high conductivity.

We can mathematically formulate the problem of steady current flow on the exterior of the battery, through a wire surrounded by air, with the differential equations $$ \vec \nabla\times\vec E =0\ \ \ \ \ \ \ \ \text{(Faraday's law)}$$ $$ \vec \nabla\cdot\vec J =\vec \nabla\cdot(\sigma\vec E)=0\ \ \ \ \ \ \ \ \text{(Conservation of charge)}$$ or conveniently with the single equation $$ \vec \nabla\cdot(\sigma\vec \nabla V)=0$$ where $V$ is the electrostatic potential, and $\sigma$ is a function of position: zero in air and non-zero in the wires. $\sigma$ contains the information about the wire geometry and is responsible for confining the current to the wire.

Puk
  • 14,100