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Why does current from a grounded electrical outlet flow through me when I touch a "hot" terminal, if stay on the floor made of a dielectric material?

Consider an AC voltage source, such as a wall socket, at $220 \, \mathrm{V}$ and $60 \, \mathrm{Hz}.$ I stay on the wood, tile, or another dielectric material, which the floor commonly consists. The floor is dry. I touch only one terminal (not ground), it hurts me. How?

I know for sure, that the screwdriver-indicator works because of resistor with huge resistance, like $2000 \, \Omega$ + body resistance, we get $$ I ~=~ \frac{220 \, \mathrm{V}}{\sim 3000 \, \Omega} ~=~70 \, \mathrm{mA} \,,$$the current flows, bulb lights, we do not die. Hence the resistance of wood is less than $2 \, \mathrm{k} \Omega \, ?$

2 Answers2

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Just wanted to clarify a couple of points.

First, even if the floor is covered by tile (good dielectric), you don't always know what is underneath it: it could be damp plywood or a pipe, which would be capacitively coupled to your body, despite the tile in-between. The current level still won't be significant, but it could be higher than expected.

Second, the effective resistance of a screwdriver-indicator should be much greater than $2000ohm$: $70mA$ current could be lethal. Below is a table copied from this OSHA site, which shows the effect of human body exposure to various current levels for $1$ second:

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V.F.
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As the mains supply is alternating current the charging and discharging of the capacitance of your body relative to free space (actually your other closer surroundings) causes a small AC current to flow and will illuminate the test indicator.

You are feeling this real current because if flows into you from a single point.

Also even a poor conductor like rubber and wood will still conduct a bit.

KalleMP
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