# Questions And Answers

# |
List All Questions | Search | List by Category |

^{2}. The logic behind this can be explained using field theory, but a simpler explanation can be used when simplifying field theory into lumped circuit theory.The painted conductive surface becomes a huge capacitor, similar to a parallel plate capacitor. The simplified equation for capacitance in this situation is the capacitance (C) is equal to the area (A) multiplied by the dielectric constant (e) all divided by the spacing between the substrate and paint (d). C = eA/d

Let’s assume that the dielectric constant of the material is unity (1) for this case (which is the approximate dielectric constant for air....so the total dielectric constant is just 8.85x10

^{-12}F/m) and the spacing is 2 mils, so the capacitance for a 20 ft^{2}Statproof® painted surface is 6,474 microfarads. Typical human body capacitance is 50 to 150 picofarads or about 65,000 times smaller than the painted floor. So a charge residing on the human body (say 5,000 volts) will be neutralized (sucked away) in record time (much less than a second ~ 0.01 seconds) because the floors capacitance is HUGE in comparison to the human body or other charged objects that may come into contact with the floor.The following analogy may prove useful........imaging that the painted floor is Lake Erie and the charged human is a glass of water. First measure the depth of Lake Erie, then pour the glass of water into it, now measure it again......see any changes? Probably not, cause the Lake is so HUGE in comparison to the glass of water. Same with the floor accepting the charge imbalance of the human body, after neutralizing the body, the floor still appears at zero volts.....or virtual ground.Second, for 800 meters square, this is about 8611 feet square and the paint averages 282.5 feet square per mil thickness, so we recommend a minimum of 61 gallons of Statproof® paint to cover your floor. If you have any more questions, please feel free to contact me.
Related Categories:

If you have found this Q/A useful, please rate it based on its helpfulness.

This question has been rated:

(0% at 0 Ratings)