Charge on fingertip during 5 mm spark to metal truck

[Josh123]

here's a problem that I am currently working on. I know the theory, but I'm not quite sure how to start this particular problem. Here are the values given:

"Thomas was crawling around on the rug. When he reached for a metal truck, a prominent spark lasting 5mmsec appeared between his fingertip and the object. His fingertip was about 2 mm from his toy. His finger burnt (the area of the burned region was of 10^-4 m^2)

On that day, the air was cold and dry causing it to become conducting when the electric field reached 3*10^6 N/C."

My question is, how you I determine the charge on the child's fingertip?
How do I estimate the resistance of the dry air between the toy
truck and the child's fingertip? (I just would like to know how to start this problem.. you don't have to do the entire thing)

Thank you in advance

 

[whozum]

Well you know the field reached 3*10^6N/C, and you know the radius (2mm). I don't know what "5mmsec" is though. I can't think of anything beyond that.

 

[Josh123]

I'm about at the same place as you. I also know that 3*10^6N/C = 3*10^6 V/m ... but I have no idea what to do with the area or even the distance

 

[whozum]

$$q_1q_2 = \frac{Fr^2}{k}$$, and your unit conversion won't get you anywhere.

What unit are you covering in class?

 

[Josh123]

It's not exactly a unit. It's a comprehensive assessment that combines several units.

 

[whozum]

If you can find the charge, then you can find the current by the relationship between charge and current. With that and ohm's law I would imagine you can find the resistance.

 

[Josh123]

The thing is, to find the charge, I need the force... to find the force, I need "k" which I don't know

 

[whozum]

$$k = \frac{1}{4pi\epsilon_{air}}$$

How would you find the charge with that?

 

[Josh123]

If Eo = 8.85 x 10-12 C2/N-m2, then k= 9.0 * 10^9

I know that E=kq/(r^2)

meaning q=1*10^-9 C

Make sense?

 

[Josh123]

If what I did in reply #9 correct:

now that I have the charge: 1*10^-9
and the time intervale 0.005 sec

then the current should be 0.0000002 A

Can you tell me if I am on the right track?

 

[whozum]

I guess #9 makes sense. Then the current would be

$$\frac{10^{-9}/}{(5x10^{-6})} \mbox{ which should be } 0.2 x 10^{-3}$$ assuming that "5mmsec" = 5 microseconds.

 

[Josh123]

I think I should be ok now and I'll be able to work out the other 10 questions associated to that problem. Thank you so much for your help

 

[whozum]

You should probably wait around for a more complete and thorough answer.

 

[Josh123]

ok.. You mean I should wait for someone else?

 

[whozum]

Yeah, also that dielectric should be a little less than 8.85x10^-12, since it is in air, so your k value should be a bit less.

 

[Josh123]

By the way, you wouldn't happen to know if there is a purpose for the area:

"the area of the burned region was of 10^-4 m^2"

 

[whozum]

Yeah Claude Mille replied to your post in the General Physics forum, check it out.

Please avoid posting in more than one forum.