Measuring Electric Fields w/Neon?

  • #1
Clyde
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Good day everyone! n00bie here.

I did do a search before I posted this... found a lot of info, most of which I couldn't apply to what I am doing (because I wouldn't know how).

I have something that I am working on that creates an electric field to alter certain chemical compounds. The distance between the two electrodes is around 1.5". I am trying to measure, or at least, get an idea, of what the field strength is inside the electric field (IE: how many v/cm). I don't need an exact measurement, just need to confirm that I am at or above a certain point.

Someone suggested a method using small neon bulbs (I think he said neon). He suggested two wires that would then be placed inside the field. Along with a few other things he would add to this 'meter', would allow the bulb to glow if the field was at a certain strength.

Any ideas on how I can measure the voltage inside this field?

Thanks!
 

Answers and Replies

  • #2
Drakkith
Staff Emeritus
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What's the voltage between the 2 electrodes?
 
  • #3
Clyde
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What's the voltage between the 2 electrodes?

i built a custom power supply that operates up to 2000v RMS. The electrodes are a little over 1.25" apart. 600-900v maybe?

Part of this project is to understand how to apply the electric field so that the v/cm is anywhere from 200 - 800v. Having an impossible time measuring it which is where this thread came from.

I'm no EE, so pardon if I'm using incorrect terminology or phrasing this awkwardly.
 
  • #4
Baluncore
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The problem with using a neon discharge to measure the gradient is that it will change the gradient by clamping the electric field. You would need a high impedance sensor to accurately measure the AC electric field.

600VAC to 900VAC. Is it a square or a sinewave? If your supply generates a sinewave then you can use RMS in the computations, otherwise the PP voltage will need to be considered.

Why do you need to measure the electric field strength when it can be computed from the VAC excitation along with plate separation and maybe geometry?
 
  • #5
Clyde
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0
I was taking the voltage setting and dividing it by the number of centimeters between the plates. It's a sine wave.

I'm going by what my EE told me to do to calculate the v/cm. I have a hard time following him completely sometimes because I don't have an electrical engineer degree.

I want to measure the field to make sure that when I am working my tests, I know for sure the field strength is where it should be. I don't have a sting enough voltmeter to measure it unfortunately, otherwise, I'd do that and just do the math.

Am I missing something? Should I be Approaching this a different way?
 
  • #6
Baluncore
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I'll make an assumption that your sine wave is symmetrical about ground, so the plate voltages are always equal magnitude but opposite polarity. Measuring the voltage on one plate relative to ground will give you less than 1kV RMS = +/–1415V peak relative to ground.

Now consider attenuating that +/–1415V with a chain of ten resistors to ground, 100k each with a 100pF, 200V capacitor in parallel. That will give you +/–150V peak AC across the lower resistor with an impedance of less than 100 k. You can measure that with a cheap digital multimeter on the AC voltage range. Multiply the meter RMS by 20 and you have your RMS plate differential voltage

The parallel capacitors make the AC response of the resistive attenuator independent of frequency and less sensitive to cheap meter capacitance.
 

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