Electrostatic induction in a conductor

[grantc]

Two aluminium plates, each 150 x 150mm, were separated by 100mm to create a form of parallel-plate capacitor. A DC voltage of 300 volts was connected across the plates to create a (fairly) linear electric field of approximately 3000 V/m between them.
An insulated conductor was bent into a U shape with squared-off corners, with the bottom of the U approximately 95mm long and each leg of the U approximately 300mm long. This was placed between the abovementioned plates so that the bottom of the U was perpendicular to the centres of the plates, with the legs of the U adjacent and parallel to the inner surfaces of the plates.
Through electrostatic induction it was expected that about 300 volts would be measured across the legs of the U. However, there was no voltage seen, both on an oscilloscope and on a Fluke meter.
Similarly, when the plates were connected to a 50 Hz, 220 volt rms supply, there was no measurable voltage across the legs of the U.
What is the fallacy in my reasoning, please?

 

[0xDEADBEEF]

A picture would be helpful. Please clarify what you mean by DC induction. You might mean some kind of polarization, but otherwise you need a changing magnetic field to induce voltage.

 

[Born2bwire]

A picture would be helpful. Please clarify what you mean by DC induction. You might mean some kind of polarization, but otherwise you need a changing magnetic field to induce voltage.

That was my thinking as well but I figured I would wait until there was a picture up. But since the electric field is static and the object in question a conductor, I do not figure seeing any induced voltage.

 

[elect_eng]

What is the fallacy in my reasoning, please?

It's hard to be sure I've interpreted your setup correctly, but it sounds like the U shaped wire is a conductor that will look like a short and not have any voltage. In order to generate electro motive force (EMF) in a conductor, consider Faraday's Law. EMF around a loop is equal to the negative rate of change of magnetic flux in the area enclosed by a loop. Even then, you would want a full loop (or multiple loops) not a U-shape (well, actually your meter/scope leads will close the loop). But, basically, your problem is that electro-statics, can't induce a net voltage in a wire.

Perhaps, you can do the experiment, then cut the wire in the center of the U-bottom. Then each half of the U will have equal and opposite charge, just like the capacitor. Then you should see a voltage.

 

[grantc]

Many thanks, oxDEADBEEF and Born2bwire. A drawing has (hopefully) been included as an attachment.
A conductor placed in an electric field E will become polarised. This polarisation (or induced) voltage should be (E x length of conductor parallel to the field). This is the voltage I am trying to measure.
Thanks to elect-eng for the wire-cutting suggestion. I tried this fairly hazardous procedure, but as soon as the wire was cut all that I measured was noise.

 

[elect_eng]

Thanks to elect-eng for the wire-cutting suggestion. I tried this fairly hazardous procedure, but as soon as the wire was cut all that I measured was noise.

Yes, my mistake. Sorry for putting you in danger for no good reason. I'm glad you were careful.

I calculate that the capacitance is only 2 pF for your outer plates. The capacitance of the inner wire will be less (after you cut the wire), most likely. This means that any charge is going to dissipate quickly when you hook up a meter or scope. I don't know the impedance of the Fluke meter but assume it's 1 GOhm. The time constant would be of the order of a millisecond. For a scope the impedance is probably 1 MOhm (or maybe 10MOhm depending on the probe) in which case the time constant is only 1 microsecond (10 microseconds). You would need to capture the decaying pulse on the Oscope to see it and the fluke meter will not register a decaying pulse.

I need to think about this more because the discharge idea is really valid if you remove the inner wire before taking the measurement. If the wire stays inside, it's less clear to me.

 

[elect_eng]

Forgetting about my wire cutting idea, I'm going to try a different explanation for your negative results.

You are placing the inner conductor within a capacitor in order to induce a charge polarization. Negative charges will be pulled near the positive plate, leaving positve net charge near the negative plate. These charges move because the inner wire is a conductor, and if you pull the wire out, the charges will flow back to neutrality.

Now consider the way you measure voltage with a meter. The meter has high impedeance which allows a small (hopefully negligible) amount of current to flow. The measured current in this resistor is an indicator of the voltage present. However, you already have a condition in which the charges can't flow back because of the outer capacitor. If they can't flow back through a low impedance conductor, then they surely won't flow back through the 1 GOhm resistance in the meter. Hence the meter sees no current in the internal resistance and concludes that there is no voltage.

Returning to the wire cutting idea, no current flows in the meter here for the same reason. There is no longer the shorted conductor inside, but we already know that the charges want to stay where they are.

This is an intuitive way of thinking in terms of circuit theory. Looking from the point of view of field theory, we can still go back to Faraday's Law. There is no changing flux, so the line integral of electric field around any closed path is zero. Of course, you want to say that the interior has electric field. However, if you charge the outer capacitor and develop induced charge difference on the inner wire, you create a zone of zero electric field (I'm being very approximate here just to get the idea across) in a region of the capacitor interior (charge shielding). However, if you then discharge the outer capacitor (or pull the wire out quickly), you will mometarily have a voltage, but it will quickly decay as the charges go back to the neutral state.

Anyway, I'm just putting some ideas out to think about. This is actually an interesting situation and I'll think about it some more.

 

[grantc]

Hi elect-eng.

My first reply to you disappeared into the message black-hole. I'm still learning how to use the Forum.

However, you already have a condition in which the charges can't flow back because of the outer capacitor. If they can't flow back through a low impedance conductor, then they surely won't flow back through the 1 GOhm resistance in the meter.

True, the charges in the conductor are prevented from flowing back because of the outer capacitor. However, any flow through the meter resistance is in the opposite direction. If you look at the attachment diagram, rmoving the external capacitor will cause the electrons in the conductor to move DOWN. If a meter is connected to the wires, the path for the electrons will be through the meter and back UP through the conductor.

While this flow is taking place, there will still be unbalanced charges across the conductor, which will give the voltage to be measured.

The interior of the conductor has a NET zero field, since inside the conductor the polarisation field will cancel the external field. This can only remain zero while the polarisation charges are present. Any flow of electrons away from the conductor and through the meter, will immediately cause the polarisation electrons to be replaced by electrons from the conductor.

The conductor in this case should behave exactly as a similar "shorted" conductor in a changing magnetic field and act as a source of voltage.

Unfortunately, the only place this seems to work is in the mythical town called "Theory", where everything works.