1. Apr 20, 2005

### 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)

2. Apr 20, 2005

### whozum

Well you know the field reached 3*10^6N/C, and you know the radius (2mm). I dont know what "5mmsec" is though. I cant think of anything beyond that.

3. Apr 20, 2005

### 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

4. Apr 20, 2005

### whozum

$$q_1q_2 = \frac{Fr^2}{k}$$, and your unit conversion wont get you anywhere.

What unit are you covering in class?

5. Apr 20, 2005

### Josh123

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

6. Apr 20, 2005

### 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.

7. Apr 20, 2005

### Josh123

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

8. Apr 20, 2005

### whozum

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

How would you find the charge with that?

9. Apr 20, 2005

### 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?

10. Apr 20, 2005

### 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?

11. Apr 20, 2005

### 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.

12. Apr 20, 2005

### 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

13. Apr 20, 2005

### whozum

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

14. Apr 20, 2005

### Josh123

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

15. Apr 20, 2005

### 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.

16. Apr 20, 2005

### 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"

17. Apr 20, 2005

### whozum

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

Please avoid posting in more than one forum.