Sorry, I fail to see why the voltage drop is 3kv across the acrylic layer. Why
is that exactly?
> Date: Tue, 3 Jul 2012 21:49:25 -0500
> To: vortex-l@eskimo.com; vortex-l@eskimo.com
> From: a...@lomaxdesign.com
> Subject: RE: [Vo]:SPAWAR has yet to respond re simple error in claims of
> effects of external high voltage dc fields inside a conducting electrolyte:
> Rich Murray 2012.03.01 2012.07.02
>
> At 08:02 PM 7/3/2012, Finlay MacNab wrote:
> >To clarify:
> >
> >An electrolyte does not conduct. Chemical reactions occur at the
> >electrodes that accept and give up electrons. Current flows through
> >the metal conductors between the anode and cathode.
>
> An electrolyte does conduct. That is, there is movement of charge.
> That is all that conduction means. Finlay, you are not being careful.
> I suggest you try it.
>
> >When I say that the voltage drop occurs withing around 1nm of the
> >electrode (the debye length), that is only the case for low voltage
> >experiments on the order of the red-ox potentials for a given
> >electrochemical reaction.
>
> Sure. The experiment is a low voltage experiment, by the way. The
> palladium deposition in this work is often done below the potential
> at which heavy water will evolve deuterium at the cathode.
>
> > At 6kV this would not necessarily be true.
>
> You aren't kidding. The thing would explode if somehow you maintained
> 6 KV across the cell.
>
> > Because the ions in the electrolyte of much much lower mobility
> > than electrons in a metal conductor they may not be able to
> > effectively screen the high applied fields, especially if the
> > solution is being mixed (a quick search of the literature did not
> > yield a relevant example at high field) . If the fields were
> > oscillating, the E field would definitely be felt within the
> > electrolyte (this is what I would have done).
>
> Well, there is work with oscillating fields, but they are oscillating
> the electrolytic current.
>
> You seem to have a concept of an electric field that is different
> from how such fields are understood by electronic engineers and
> physicists. You ask a question below that is actually quite easy to answer.
>
>
> >When you say:
> >
> >The situation has nothing to do with "free charges that can migrate
> >to the surface" of anything. The mode of conduction is irrelevant.
>
> There is no surface here, not that is defined. There is a conductive
> medium, it has a certain resistance. Current flows through it when
> there is a potential across it (actually such electrolytes can also
> generate potentials, I won't go there.). Ohms law is obeyed.
>
> >I fail to understand what you mean. The only reason that the field
> >inside the electrolyte can be zero is if charge carriers migrate to
> >the surface of the cell to screen the bulk of the electrolyte from
> >the externally applied field.
>
> No, any region of low potential "screens" the field. You are making
> it much more complicated than it is. Imagine a line between the two
> high-voltage plates. Imagine an equipotential region inside the
> electrolyte, parallel to the plate, the line crosses that region.
> Let's assume, to keep this simple, that the equipotential region is
> larger than the high voltage plate. How can the high voltage on the
> other side of this equipoential region affect *any* region beyond the
> equipotential region?
>
> This is DC, remember. There is a very high voltage drop across the
> acrylic, about 3 KV. That's a done deal! The voltage doesn't then rise up!
>
> >I don't believe your example of probing the electrolyte with two
> >probes and a bridge is relevant to this experiment, since the
> >external electrodes are not in contact with the electrolyte, no
> >chemical reaction can take place, and so no current can flow, the
> >field can only be screened by the build up of charged ions at the cell walls.
>
> This is how to measure voltage! Because no charge movement is
> involved, the whole issue about charge carriers is irrelevant. That's
> why a bridge is used, in fact. In practice, what is needed is a very
> sensitive current meter, which is zeroed out by applying the
> reference voltage and adjusting it until the current is zero.
>
> A "bridge" here just means that current is measured, and the
> experimental voltage is measured by opposing it with a known voltage,
> such that no current flows in the circuit.
>
> >Maybe you can explain it in a way i can understand.
>
> Sure, I hope.
>
> >What would happen if you had two metal plates separated by an air
> >gap, then you applied a 6kv bias between them, and then put your two
> >probes into the air gap?
>
> Air conducts electricity. If the air conducts uniformly, the
> resistance of the air will be even and the voltage will uniformly
> decline, linearly, between the plates. The bridge will measure the
> voltage accordingly.
>
> In the subject example, there are three regions between the plates.
> Two regions are filled with acrylic, which has very high resistance,
> higher than air, if I'm correct. And then there is the electrolyte,
> which has relatively low resistance.
>
> The voltage gradient is directly proportional to the current times
> the resistance per unit length.
>
> That's simply another version of Ohm's law.
>
> In an elecric circuit, we do not need to know what voltages are
> present elsewhere in the circuit to know the relationship between
> current, voltage, and resistance, in one leg of the circuit. Electric
> field strength is just another name for voltage gradient.
>
