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

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Homework Help Overview

The problem involves calculating the total electric charge discharged from a lightning rod over a period of time and determining the corresponding number of electrons involved in the process. The context is set during a thunderstorm, focusing on the charge flow from the rod into the surrounding air.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the total charge by multiplying the charge per second by the number of seconds in an hour but expresses confusion about the accuracy of their results. They mention a consistent discrepancy by a factor of approximately 0.016, leading to questions about their calculations.

Discussion Status

Participants are engaged in clarifying the calculations needed to find the total charge and the number of electrons. The original poster has acknowledged a potential misunderstanding regarding the number of seconds in an hour, which may have contributed to their errors. There is no explicit consensus yet on the correct approach, but some guidance has been provided regarding the calculations.

Contextual Notes

The original poster is working under the constraint of having limited attempts to submit their answers correctly, which adds pressure to their calculations. There is an indication of confusion regarding basic units of time and their impact on the final results.

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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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 ?
 
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~
 
No prob. :smile:
 

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