-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?

tryst with freedom
  • 7,940

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
  • 13,526
  • 1
  • 22
  • 42
  • Feel like you missed the question – tryst with freedom May 13 '23 at 18:31
  • 2
    @TrystwithFreedom The questions I see are "does E-field exist only in the wire?" and "how do we justify current flows in wires?", both of which I tried to address. – Puk May 13 '23 at 18:33
  • Right, but the equations you wrote down arise from theory of matter involved in E&M (as discussed in linked post). My question is, essentially, how is this fact justified in the theories of matter? – tryst with freedom May 13 '23 at 18:34
  • @TrystwithFreedom Are you asking why Ohm's law is valid? – Puk May 13 '23 at 18:35
  • That I understood, it's just an experimental observation (as is explained in a linked post in the post I linked). I am asking if there is a more fundamental reason for eletric field being mostly confined to the wire. – tryst with freedom May 13 '23 at 18:37
  • @TrystwithFreedom I don't understand. The relevant equations are Ohm's law and Maxwell's equations, which are what I used. Are you asking for something more fundamental than these? – Puk May 13 '23 at 18:39
  • 3
    @TrystwithFreedom, the electric field is not confined to the wire. But outside the wire there are no free carriers to flow and produce current. Puk pointed out that the conductivity of the wire is much higher than the conductivity of air, which is another way to say the same thing. – The Photon May 13 '23 at 18:55
  • 1
    @TrystwithFreedom Remember, physics is experimental science. Theories do not justify experimental results: nature doesn't care what your theories are. Ask instead how the experimental results lead to the theory you are interested in. – John Doty May 13 '23 at 18:59
  • 1
    I would have just drawn a circuit diagram, battery connected to resistive load, and resistive load is parallel to resistor labelled as air resistance. Since the air resistance is huge compared to typical loads connected by wires, the electric current chooses the path of least resistance and so the current passing through air is a tiny contribution compared to the current passing through wires. – naturallyInconsistent May 14 '23 at 14:17