How does the amount of charge on a doorknob affect the length of a spark?

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The discussion revolves around the relationship between the charge built up from scuffing feet on a carpet and the resulting spark when touching a doorknob. It highlights the breakdown of air at a field strength of 3.0 x 10^6 N/C and the calculation of charge using the formula Q = E(d + r)^2/kc. Participants express confusion over the lack of a specific value for the constant Kc and how to incorporate the doorknob's charge into their calculations. Additionally, there is a query about the electron transfer process during scuffing and whether it can occur in reverse. The conversation emphasizes the need for clarity in the equations and constants used in electrostatic calculations.
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Homework Statement


On a dry winter day, if you scuff your feet across a carpet, you build up a charge and get a shock when you touch a metal doorknob. In a dark room you can actually see a spark about 5.0 cm long. Air breaks down at a field strength of 3.0 x 10^6 N/C. Assume that just before the spark occurs, all the charge is in your finger, drawn there by induction due to the proximity of the doorknob. Approximate your fingertip as a sphere of diameter 1.5 cm, and assume that there is an equal amount of charge on the doorknob 5.0 cm away.
How much charge have you built up?

Homework Equations


Q*Kc/r^2

The Attempt at a Solution


E > or equal 3*10^6
Knob)----d---(---R
2cm
(E*(d+r)^2)/kc=Q

Now I have several problems with this, first there's no value of Kc explicitly given to me. Our professor explicitly gives us Kc when it needs to be used to a certain decimal place. Secondly, I'm having an issue figuring out what exactly to do with my charge from the doorknob. My instinct tells me that both will be contributing to the charge equally but I can't seen to figure out where the distance of it and its charge would come into play with this problem. I'm thinking it has something with Kc not being used at all near the end solution but I'm not sure.

Any advice is appreciated, thank you.
 
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The 3*106 N/C is also 3*106 V/m

At 5cm then you know that the potential difference - thumb to knob must be 1/20th of that.

I'd simply use 8.99 * 109 for k in air.
 
Tried with 8e9 and got it returned as wrong so hrmm, still scratching away at this..

I've tried also to do this

Q=E(d+r)^2/kc (finger to any point where air breaks down)
Q=E(r)^2/kc (doorknob to any point where air breaks down)

Soo:

E(r)^2/kc=Q=E(d+r)^2/kc
But this ends up giving me units of meters alone so.. that won't work either
 
quick question: why is it when you scuff your feet, you lose electrons, instead of gain electrons from the carpet? Is it possible for it to go the other way?
 
Specialmias said:
On a dry winter day, if you scuff your feet across a carpet, you build up a charge and get a shock when you touch a metal doorknob. In a dark room you can actually see a spark about 5.0 cm long.
Surely they mean 5 mm here? :bugeye:
 
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