# Time-Varying Electric Field in a Wire

1. Sep 2, 2011

### kmarinas86

When, $dE/E >> 1$, isn't it possible to have a time-varying electric field that produces a displacement current greater than the conduction current?

Would it also be the case that one could achieve this effect by switching a circuit on for a small period of time, say where $T << 1/f_0$?

What happens if a person were to switch a circuit on quickly, but switch it off much quicker? It would cause a massive back-spike of displacement current in the opposite direction. If there was an inductance, there might also be some resistance to changing it before and after it was established. If the E-field was produced by discharging a capacitor into an inductor, what happens to the back-spike of displacement current, generated by disconnecting the inductor from the capacitor, once it arrives at the capacitor?

Does displacement current (provided it has voltage) transfer reactive power or real power? Is there a way to quantify the power conveyed through the displacement current, and if so, what is it equal to?

And finally, what energy source does displacement current draw upon to exist in the first place, when created by shorting a capacitor, given that it is not proportional to the movement of free electrons away from the capacitor under various situations satisfying the conditions $dE/E >> 1$ and $T << 1/f_0$?

Last edited: Sep 2, 2011
2. Sep 3, 2011

### Staff: Mentor

For a lumped element circuit the displacement current is always exactly equal to the conduction current. Otherwise the capacitor would violate KCL.

3. Sep 3, 2011

### kmarinas86

It is of course clear in a closed loop circuit that the conduction current going into and out of a capacitor should equal the displacement current going through the capacitor.
However, I was thinking more on the lines of a wire.

I imagined that displacement current and conduction current would both contribute to the total current in a wire when $dE/E >> 1$.

I am considering a scenario where the wire has a parasitic capacitance, thus implying that is possible to have both displacement current and conduction current in the same wire element.

4. Sep 3, 2011

### Staff: Mentor

What is dE/E? That isn't good math, so I don't know what you are trying to say.

5. Sep 3, 2011

### kmarinas86

I meant the change of E per E with respect to time. To spell it out, this is $(dE/dt)/E$ (where $dt$ is set to unity, i.e. $=1$). If E is small and E increases rapidly, then $(dE/dt)/E$ is large. If E is large and E is not increasing very fast, then $(dE/dt)/E$ is small.

If $(dE/dt)/E$ is constant with respect to $t$, $E(t)$ follows an exponential function. $dE/E$ is basically the % change in E.

Last edited: Sep 3, 2011
6. Sep 3, 2011

### Staff: Mentor

Well, you can always make that quantity much larger than 1 simply by your choice of units.

7. Sep 4, 2011

### kmarinas86

True, but what about $T << 1/f_0$? Certainly it's not the case for that.

8. Sep 4, 2011

### Staff: Mentor

What is f0? Is the circuit you are talking about switching a specific kind of circuit?

9. Sep 4, 2011

### kmarinas86

The resonance frequency of this specific circuit, which is a wire connected in series with a discharging capacitor. We could also say the wire could be wrapped as a solenoid to give it significant inductance (rather than some tiny stray inductance), reducing $f_0$ further yet. For there to be displacement current in the wire itself, it would also be required that it have some stray capacitance as well.

According to classical EM theory, the displacement current also generates a magnetic field. However, I never have seen anyone try to demonstrate this in a plain wire or a solenoid. If the wire has a stray capacitance, you would think that perhaps there would also exist a magnetic field due not just to conduction current in the wire but also the displacement current in the wire, existing in tandem with it. Of course, this would be very small in normal situations, but because the displacement current is proportional to the rate change in $E$, whereas the conduction current is proportional to $E$, I think there should be some special situations where the reverse is true. This would mean it is possible to generate a magnetic field in a wire that is much more than what you would expect from the conduction current alone.

The problem is, I thought the discharging of a capacitor was proportional to conduction current. I don't see how displacement current by itself would discharge the capacitor. So what I also wanted to know is what is the energy source that is withdrawn to generate the displacement current in a wire.

Last edited: Sep 4, 2011
10. Sep 4, 2011

### Staff: Mentor

Certainly you can switch a circuit on or off faster than the resonant frequency. Depending on the details of the circuit and switch that can lead to arcing.

I don't know what you are getting at here. In EM the displacement current is often greater than the conduction current. In fact, in free space it is always greater by definition. In circuit theory they are always the same, by the assumptions of circuit theory. All it means is that you have to use Maxwells equations to analyze such a situation, and not circuit theory.

11. Sep 4, 2011

### kmarinas86

When I switch on the voltage in an inductor, the current lags behind it. But normally this is considered to only be conduction current. The relevant decay constant for the current would be L/R, which is inductance divided by resistance, while the relevant decay constant for voltage would be R*C, which is resistance times capacitance. The larger the decay constant, the longer time involved in charging and discharging with voltage and/or current. So if we wanted the conduction current to be smaller, one should aim for a smaller V/R, a higher L/R, and a smaller time interval T, and if we want the voltage to increase rapidly, one should aim for a higher V and a smaller R*C.

The SI unit for the electric field is volts per meter, so it would seem to me that for a given device, the rate of increase of voltage, divided by the length of the wire, is proportional to the rate of increase of the E-field. To be more precise, the displacement current is proportional to the rate change in electric flux, which is given by the integral of the E-field with respect to an area. So the factor of cross-sectional area of the wire divided by length, when multiplied with the rate change of the voltage (i.e. V/s), gives us a figure that is proportional to the displacement current.

So, for a given V and an arbitrarily small T, we have a conduction current that is proportional to (1/R)*(R/L) or 1/L and a displacement current proportional to (wire cross-sectional area/wire length)/(R*C) or (resistivity)/(resistance^2*capacitance). When our resistivity is a given by the material we chose, the ratio of displacement current to conduction current, when T is arbitrarily small, varies proportionally to L/(CR^2). Ignoring any losses, this is equal to the square of the ratio of the characteristic impedance to the DC resistance. If we decide to consider the effect that the winding diameters and inter-winding spacing has on the L/C term, then, if ignoring losses, the L/C factor varies proportionally to the square of the log of the ratio of the distance between winding centers and the radius of the conductor and varies inversely with the permittivity of the insulation material. So in other words, this effect, for a wire of given dimensions, if my calculations are correct, would improve with increased insulation thickness. So, to maximize the ratio of displacement current to conduction current, one might make a solenoid (or even better, a Brooks coil) out of wire consisting of very thick insulation, which is, ironically, totally opposite of what is usually desired in a solenoid. Appropriately enough, it seems that insulation is what is needed to prevent the arcing which you mention. Perhaps when such a coil is constructed, wherein most of the field is produced by displacement current produced shortly after a capacitor is discharged across the coil, certain questions should be raised as to what source of energy is the displacement current in a wire withdrawn from, as it does not require the displacement of electrons out of the capacitor.

12. Sep 4, 2011

### Staff: Mentor

So this whole thing is actually a conservation of energy question?

http://farside.ph.utexas.edu/teaching/em/lectures/node89.html