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How to count coulombs?

  1. Mar 12, 2009 #1
    What is the best way to put a charge on a plate?
    By 'plate' I mean some mass of conductor surrounded by an insulator. The problem, really, isn't how to generate charge, I know I can use a Van de graff or high-voltage rectifier for that; the problem is that I need to be certain of how much charge is on the plate.

    I've looked into using a noncontacting field meter, but if I understand things right, I also need the capacitance to determine the charge on the plate. Also, I know that if I could precisely measure the current as the plate was being charged, I could combine that with the time to get the coulombs, but that is such a short time and such a small current that I doubt it would be very effective.

    Any suggestions would be greatly appreciated.
     
  2. jcsd
  3. Mar 12, 2009 #2

    clem

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    A "ballistic galvanometer" measures the total charge passing through in a short time.
     
  4. Mar 12, 2009 #3
    If you charge the plate through a series large known capacitor C (intially discharged), and then measure the voltage V across the known capacitor, the charge on the known capacitor C*V is the same as the charge on the plate.
     
  5. Mar 13, 2009 #4
    Is it really that simple?!
    That's because of Kirchoff's circuit equations (eg, constant current in series), right?

    So you're saying something like
    http://rprogrammer.net/static/curios/Monoplate_charger.gif [Broken]
    or
    http://rprogrammer.net/static/curios/Biplate_charger.gif [Broken]

    The first one would be more convenient for me, but I imagine I could get better charge on one plate by charging two, then physically separating them afterward.
     
    Last edited by a moderator: May 4, 2017
  6. Mar 13, 2009 #5
    The second one has a complete circuit loop. You will need to ground one plate. If your "plate" is a free-standing object in air, then the object has capacitance to ground through the air to the room. So the physical ckt looks like the first one, and the return loop still exists through the room ground. Just follow the field lines. Be sure to ground one side of your high voltage supply.

    Do you want to charge the two plates with opposite polarity? If you charge two plates with a separation d, and then pull them apart, you reduce the capacitance and increase the voltage. Remember Q = C*V, and Q is conserved.
     
    Last edited by a moderator: May 4, 2017
  7. Mar 13, 2009 #6
    I know that the single-plate one logically connects to ground through the air; what I meant was: Would the capacitance it could have being that far from a ground be at all significant?

    My current setup requires a single plate with one charge (the polarity doesn't matter, although I hear positive is less leaky in general). If I used two plates in the charging process, I would have to separate one plate and discharge it elsewhere.
     
  8. Mar 13, 2009 #7
    Also, wouldn't the control capacitor leak after I disconnected it to measure its voltage?
     
  9. Mar 15, 2009 #8
    The capacitance of an "isolated" plate to the nearby walls is what establishes the relationship between charge and voltage on the plate; e.g., C = Q/V. If you have net charge on a plate, then there have to be field lines near the plate. They have to terminate someplace on an opposite charge. You cannot have dangling field lines.

    I have made special low leakage voltage amplifiers using low leakage capacitors that have held their charge for hours. Mica caps are usually pretty good. An old 365 picofarad variable tuning cap in an old AM radio is very good. What you need is an electrostatic voltmeter, or a good opamp in a voltage follower configuration (with your measuring cap across the inputs). The Analog Devices AD8605 or similar CMOS op amp with a 1 picoamp input offset current sells in quantilies for 68 cents. If your plate has 1 picoFarad capacitance to the environment, and you put 10kV on it, then you have 10^-8 Coulombs on it. Then your series cap should be about 0.001 to 0.01 microfarads for a starter. If you have 10^4 picoCoulombs on a capacitor and a 1 picoamp leakage rate in the opamp, then the discharge time constant is about 10^4 seconds.
     
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