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

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.

BvU
Homework Helper
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.

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...

BvU
Homework Helper
I used 1st result in Google "Surface area of hand" so no I only used one plate
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 ?

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..

???

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 ?

smokingwheels
berkeman
Mentor
Looks as if teacher should be in jail -- still.
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.

smokingwheels