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Electrostatic discharge and metal rods

  1. Aug 1, 2014 #1
    I uploaded a photo to help explain the question, basically the blue rod is a -ve charged rod ( high voltage) gray one is metal and the brown is also metal but it's grounded , if gap A and B are filled with air , the high voltage will break down the air in both gaps and make 2 sparks in both gaps , but what if gap A was filled with water ( or a liquid with lower dielectric constant) , gap B with air, and the voltage is not high enough break down the water , can the charged rod induce a charge on the metal rod in the middle high enough to break the air in gap B making a spark in gap B only , leaving the middle metal rod +ve charged
     

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  3. Aug 2, 2014 #2

    NascentOxygen

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    The water's conductivity will probably allow it to conduct well enough to cause the airgap to break down.
     
  4. Aug 3, 2014 #3
    it's pure water, if you meant something else please explain
     
  5. Aug 3, 2014 #4

    NascentOxygen

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    Air has εr=1, water has εr≈60-80. Is this suitable for your experiment?

    Replacing air by pure water will lower the voltage across the water gap, leaving most of the potential difference to appear across the air gap until that breaks down.
     
  6. Aug 3, 2014 #5
    but if the voltage difference was decreased because of the permittivity of the water , doesnt that mean the voltage or charge induced on the second rod will be reduced?
     
    Last edited: Aug 4, 2014
  7. Aug 3, 2014 #6

    NascentOxygen

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    You are using "electric strength" (meaning "dielectric strength", maybe?) where I expected you would be using "relative permittivity", so what property do you mean?
     
    Last edited: Aug 4, 2014
  8. Aug 4, 2014 #7
    sorry , i was on a hurry and got confused, i mean the relative permittivity, i will edit my old post and correct the error
     
  9. Aug 4, 2014 #8

    NascentOxygen

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    If the blue rod has a fixed potential, then the dielectric with higher permittivity will carry a smaller proportion of that potential until dielectric breakdown occurs, after which the intact dielectric experiences almost the full applied potential. Provided that second dielectric does not breakdown you have a capacitance charging up, and when current flow ceases the centre rod will have a nett positive charge.
     
  10. Aug 4, 2014 #9

    NascentOxygen

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    You have yet to address my inquiry in post #4.
     
  11. Aug 4, 2014 #10
    This is what i understood so far , most of the voltage will be applied to the air gap , until it breaks down, leaving the metal rod +ve charged , then all the voltage will be applied to the water gap , right?
    if you were asking about the permittivities , yes , air = 1 , water = 60-80
     
  12. Aug 4, 2014 #11

    NascentOxygen

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    Perhaps not all will be applied to the water. During breakdown of the air there will be a minimum voltage needed to maintain that conduction, this reduces the voltage to the other dielectric during this time.
     
  13. Aug 5, 2014 #12
    is that minimum voltage only required when the breakdown occurs in microseconds?
     
  14. Aug 5, 2014 #13

    NascentOxygen

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    No. It's the voltage across the spark.
     
  15. Aug 5, 2014 #14
    so what happens after the spark?
     
  16. Aug 5, 2014 #15
    E= mc2
     
  17. Aug 5, 2014 #16
    is this a spam of some kind?
     
  18. Aug 5, 2014 #17

    NascentOxygen

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    When electron flow ceases, the situation reverts to a pair of capacitances. They share the remaining potential, but differently now that one had briefly shorted out (but being a gas, that dielectric healed itself).
     
    Last edited: Aug 5, 2014
  19. Aug 6, 2014 #18
    what do you mean by "remaining voltage"?
     
  20. Aug 7, 2014 #19

    NascentOxygen

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    I took the blue rod as carrying a finite number of charges and separated from earth by capacitance and dielectrics. When the dielectrics change, such as when one shorts out, the rod's potential relative to earth changes.
     
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