How does the electric field pattern looks like when current is moving inside the conductor and why this field is uniform?

Question

We know that electric field is uniform inside a normal electric circuit consisting of some battery source and a conducting wire , $\mathrm{E}$ will be $\mathrm{E=V/d}$ , but I cannot think of a reason as to why this field is uniform and also how this field $\mathrm{E}$ looks like inside the conducting circuit and outside too? Is it different from the stationary charge electric field?

Answer 1

It's known that in the static case (i.e., $\partial/\partial t=0$) there is no electric field inside a conductor. You can prove this for yourself by solving the continuity equation $\frac{\partial\rho}{\partial t}+\nabla\cdot\vec J=0$ in a conductor where $\vec J=\sigma\vec E$. The higher the conductivity, the shorter the Maxwellian relaxation time $\frac{\epsilon_{0}}{\sigma}$ for the charge to spread on the surface of the conductor as opposed to being inside of it. Thus, there can be no electric field inside the wire and no electric potential on the two ends of the wire since the electric field on the surface of the wire will be perpendicular to the surface. However, the electrons will still drift through the surface of the wire with constant velocity as they will not accelerate or decelerate. Electrons will get the energy to circulate through the circuit when they pass the battery and they lose some energy when they pass through e.g., a resistor.

The specific relation $E=V/d$ that you provided often happens in modeling capacitors where two conducting plates are separated from each other at a distance $d$. It is an assumption that $E$ in the space between the plates is uniform, constant, and equal to $V/d$ where $V$ is the potential difference between the two plates. While the electric field may seem uniform in the interior sections of the plates, it definitely has curvature at the edges. Nonetheless, the fringing is considered negligible.