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

1. Jan 10, 2017

### smokingwheels

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. Jan 10, 2017

### BvU

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.

3. Jan 10, 2017

### smokingwheels

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...

4. Jan 10, 2017

### BvU

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 ?

5. Jan 10, 2017

### Staff: Mentor

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.