Solve Gale's Lightning Problem: Find the Charge & Electrons

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In summary, the conversation is about a problem someone is trying to solve with a limited number of tries. They are trying to calculate the amount of electric charge and number of electrons involved in a corona discharge during a thunderstorm, using a hint provided. However, they are consistently getting an answer that is off by a factor of about 0.016, and are asking for help to understand what they are doing wrong. After receiving a helpful tip, they realize their mistake and thank the person for their assistance.
  • #1
Gale
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HI, i have this problem that I've been working on, i get 5 tries to get it right, and i only have one try left, so i have to get this right.

O-16-3 Supposed that during a thunderstorm, the corona discharge from a dissipative lightning rod into the surround air amounts to 0.799 x 10-4 C of positive charge per second.
(a) If this discharge goes on more or less steadily for an hour, how much electric charge flows out of the lightning rod?
(b) How many electrons flow into the lightning rod from the surrounding air?

i submit my wrong answers, and it give a hint, which is:
The total charge is the charge per second times the number of seconds. The charge on an electron is 1.6 x 10-19 C. Divide the total charge, by the charge on an electron to find the number of electrons.

which is what i thought I've been doing. but i guess not. I multiply the .799x10-4 by 60. But its always wrong. I've had to do it with different numbers every time, and its always off by a factor of about .016. If i divide my answer by .016, i get something close to the right answer, but not close enough. I don't understand why its that far off everytime, or what I'm supposed to be doing. PLEASE HELP!

~Gale~
 
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  • #2
Gale17 said:
HI, i have this problem that I've been working on, i get 5 tries to get it right, and i only have one try left, so i have to get this right.

O-16-3 Supposed that during a thunderstorm, the corona discharge from a dissipative lightning rod into the surround air amounts to 0.799 x 10-4 C of positive charge per second.
(a) If this discharge goes on more or less steadily for an hour, how much electric charge flows out of the lightning rod?
(b) How many electrons flow into the lightning rod from the surrounding air?

i submit my wrong answers, and it give a hint, which is:
The total charge is the charge per second times the number of seconds. The charge on an electron is 1.6 x 10-19 C. Divide the total charge, by the charge on an electron to find the number of electrons.

which is what i thought I've been doing. but i guess not. I multiply the .799x10-4 by 60. But its always wrong. I've had to do it with different numbers every time, and its always off by a factor of about .016. If i divide my answer by .016, i get something close to the right answer, but not close enough. I don't understand why its that far off everytime, or what I'm supposed to be doing. PLEASE HELP!

~Gale~

How many seconds in an hour ?
 
  • #3
Oh my god... I am sooo dumb. Heh.. you that's probably it. Heh... i feel stupid... :redface: Thanks a bunch, i probably never would have realized that.

~gale~
 
  • #4
No prob. :smile:
 

1. What is Gale's Lightning Problem?

Gale's Lightning Problem is a hypothetical scenario where a lightning bolt strikes the ground. The question is about finding the charge and number of electrons involved in this event.

2. How can the charge and number of electrons be determined?

The charge and number of electrons involved in Gale's Lightning Problem can be determined using the equation Q = Ne, where Q is the charge, N is the number of electrons, and e is the elementary charge. This equation is based on the fundamental relationship between charge and electrons.

3. What factors affect the charge and number of electrons in Gale's Lightning Problem?

The main factors that affect the charge and number of electrons in Gale's Lightning Problem are the strength of the lightning bolt, the distance from the strike, and the properties of the surrounding environment, such as the conductivity of the ground.

4. How is Gale's Lightning Problem relevant to real-life situations?

Gale's Lightning Problem is a simplified version of real-life events, such as lightning strikes during a thunderstorm. Understanding the principles behind this problem can help scientists and engineers develop better lightning protection systems and improve our understanding of atmospheric electricity.

5. Are there any limitations to using the Q = Ne equation in Gale's Lightning Problem?

While the Q = Ne equation is a useful tool for solving Gale's Lightning Problem, it is based on certain assumptions and simplifications. In real-life situations, there may be other factors at play that can affect the charge and number of electrons involved in a lightning strike. Additionally, the equation may not accurately account for the effects of strong electric fields and non-uniform charge distributions.

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