# What does E/M field look like when I close a circuit?

Suppose that we have a charged capacitor with two pins: $C_+$ and $C_-$.

Suppose that we have a long wire with fixed geometry, that is already connected to the pin C+.
Let the distance along the wire be called $s$.

I closed a circuit by connecting another end of the wire to the pin $C_-$ in the moment $t=0$.

What is the general way to find electrical current in wire as a function of $s$, $t$ and geometry ?

Is there a difference between if the first end of wire was connected to the $C_+$ pin, or unconnected at all?

• – rob May 17 '14 at 1:08

This is a very complicated thing to describe fully: in general one will have to solve the full Maxwell equations, depending on the geometry. As soon as the circuit is made, electron currents begin to flow in the wire, beginning at the point where you touch the capacitor (the $C_-$ pin). The electron drift is established by an electromagnetic wave running along the wire at the speed of light $c$ towards the other terminal, and in general there will be a reflexion at that terminal, so you will have an extremely complicated system of waves bouncing back and forth along the wire until the steady state behavior is reached. There are two geometries that can be analysed approximately to give you insight. Firstly, if the wire is very long, roughly circular loop so that it doesn't cross itself or come near itself at any point aside from at the capacitor terminals, the it can be treated as a circularly symmetric waveguide and the field around it can be decomposed in to waveguide modes whose mathematical form I will add when I have more time to answer the question. But basically you have a linear superposition of modes and you find the unique superposition weights that match all the field conditions prevailing just after you have closed the circuit and you work out the circuit's behavior propagating this superposition to the other end, then working out the reflexion co-efficients from the $C_+$ pin, propagating the reflected linear superposition back along the wire, and so forth so you sum up all the bounces. The modes are lossy, so each "bounce" wave is less intense than its forerunner, and the sum of the waves slowly builds up to the "circuit analysis" model of the system, wherein the capacitor's voltage decays with a time constant $R\,C$ where $R$ is the Ohmic resistance of the loop.
For a more readily imagined scenario, we think of a long, thin loop so that $\ell \gg w$ where $\ell$ is the length of the loop and $w$ the approximately uniform distance between the conductors as below. At very short timescales, less than $w/c$, the top wire behaves as the general waveguide just describes and radiates as a superposition of single-wire waveguide modes across the width $w$ to begin interaction with the lower wire. Now the wire pair, being approximately translationally invariant, also has a system of waveguide modes, amongst them the TEM "transmission line modes". See the Wikipedia page on TEM transmission lines and also my discussion of an approximately TEM system here towards the end after the heading "Approximations to the General Description and a "Circuit" Description of the System". The radiation across the circuit's width quickly establishes the TEM mode of the pair as the dominant mode: since all the other modes are very lossy (which I'll show when I add the mathematical form for them above), and have much greater imaginary propagation constants (much lossier) than the TEM mode, this happens over a very short length of line indeed.
So now, at timescales greater than a few $w/c$ we have a single voltage step function $\theta(z-c\,t)$ (where $\theta$ is the unit step, and $z$ the axial distance along the loop propagating along the pair as shown. When it reaches the dead short at the other end, a reflected wave propagates back. When the backpropagating wave hits the capacitor, there is another reflexion and so forth, as governed by the reflexion co-efficients which the Wiki article tells you how to calculate. The modes are slightly lossy, so each "bounce" wave is a little less intense than its forerunner, and the sum of the waves slowly builds up to the "circuit analysis" model of the system, wherein the capacitor's voltage decays with a time constant $R\,C$ where $R$ is the Ohmic resistance of the loop.
Recall $$I = n q v_d A \tag{1}$$ Note that the electrical current depends only on the cross sectional area of the wire and not on its length... However if we manipulate eq.1 we find that $$I = \frac{nqsA}{t}$$ Since $v_d = s/t$ if we continue further we see that $$I = \frac{nqV}{t }$$ Where $V$ is the volume of the Wire. I don't get your second question, please be more clear.