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Mylar sheet, 1 sq foot charged to 20 KV, clap to discharge

  1. Jan 10, 2017 #1
    Please delete if not allowed.

    The teacher charged a 1 square foot sheet of mylar sheet up to approx DC 20 kv one day long ago in school via an modified EHT output of a TV.
    I had one of my mates hold it by the corners in line with my hands, I then judged and guessed and clapped it with both hands opposite sides to discharge it.
    I felt the compression thru my chest my heart seemed to sort its self out and that was 30 years ago.

    Not too sure about brain damage?
    I was never that good at basic school work anyway.

    Calculation.
    I am not sure about the contact closure time of hands (2 mS).
    Full Hand to Hand resistance unsure (1 k).
    Surface area of hand .054 m square.
    Mylar sheet dielectric constant 3.1, thick enough to withstand 20 kv.

    I have come up with 1.186 Joules or 600 W roughly discharged.
     
  2. jcsd
  3. Jan 10, 2017 #2

    BvU

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    You have big hands -- or did you add the two sides ?
    care to show your calculation ? You left out the mylar thickness
    Looks as if teacher should be in jail -- still.
     
  4. Jan 10, 2017 #3
    Lets say I have average hands with slightly longer fingers but skinner.
    I used 1st result in Google "Surface area of hand" so no I only used one plate.
    Calculation well Googled across 2 OS ..copy and past has a problem from Raspberry Pi Pixel OS to Ubuntu Thinlinc Server so cant post all the links to the online caluclators..
    Mylar thickness I did say "thick enough to withstand 20 kv." maybe I should of added with corona activity on insulation brake down.
    The Teacher did check my condition a few minutes later and I said, Yes I am fine so I guess that covers him...
     
  5. Jan 10, 2017 #4

    BvU

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    Surface area of hand is what gets wet when you submerge it. The contact area with the mylar is much smaller.

    How does teacher put charge on a mylar sheet ? Is the surface conducting somehow ? Or does he just rub it on ?


    ???

    You did something like ## Q = CV ##

    with ##C =\displaystyle {\varepsilon_0 \varepsilon_r A \over d } ##

    so, say ##C\approx \displaystyle{3.1 \ 8.9\,10^{-12} \ 0.054 \over 50\, 10^{-6} }\ = 30 ## nF

    (for a 50 ##\mu## thick sheet)

    and Q ## \approx 0.0006 ## C ?

    So that ##E = {1\over 2} CV^2\approx 6 ## J ?
     
  6. Jan 10, 2017 #5

    berkeman

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    No kidding, what an idiot.

    @smokingwheels -- You would need the thickness of the Mylar sheet and its area. I'm assuming it had metalization on both sides, and it was charged up like a capacitor. You calculate the capacitance using a simple formula that you can look up, involving the area, thickness and dielectric constant. Clapping it with your hands give you an arm-to-arm shock, which is extremely dangerous. You were lucky your heart didn't go into ventricular fibrillation. Thread is done.

    EDIT -- I see BvU helped you with the calculations. Thanks BvU. :smile:
     
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