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Wednesday, September 30, 2015

25: Problem 25.57

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INTRO:
You sometimes create a spark when you touch a doorknob after shuffling your feet on a carpet. Why? The air always has a few free electrons that have been kicked out of atoms by cosmic rays. If an electric field is present, a free electron is accelerated until it collides with an air molecule. It will transfer its kinetic energy to the molecule, then accelerate, then collide, then accelerate, collide, and so on. If the electron’s kinetic energy just before a collision is 1.80×10−18J or more, it has sufficient energy to kick an electron out of the molecule it hits. Where there was one free electron, now there are two! Each of these can then accelerate, hit a molecule, and kick out another electron. Then there will be four free electrons. In other words, as shows below, a sufficiently strong electric field causes a “chain reaction” of electron production. This is called a breakdown of the air. The current of moving electrons is what gives you the shock, and a spark is generated when the electrons recombine with the positive ions and give off excess energy as a burst of light.

Part A: 
The average distance an electron travels between collisions is 2.00 μm . What acceleration must an electron have to gain 1.80×10−18J of kinetic energy in this distance?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
a = 9.88×1017 m/s2

Part B:
What force must act on an electron to give it the acceleration found in part A?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
F = 9.00×10-13 N

Part C:
What strength electric field will exert this much force on an electron? This is the breakdown field strength. 
Note: The measured breakdown field strength is a little less than your calculated value because our model of the process is a bit too simple. Even so, your calculated value is close.
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
E = 5.63×106 N/C

Part D:
Suppose a free electron in air is 0.700 cm away from a point charge. What minimum charge must this point charge have to cause a breakdown of the air and create a spark? Assume the electron is not hitting the charge particle but a neutral air atom that is still 2.00 μm away.
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
q = 30.6 nC

12 comments:

  1. Replies
    1. You have to take a look at the amount of joules she used compared to what is used now, in her example 1.80×10−18J is used, but in problems used today, 2.0×10−18J is used therefore different answers are given with that simple change

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  2. How do you get any of these answers?? There's no work...

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    Replies
    1. For part a it is (1.8*10^-18)/(9.11*10^-31 * 2.0*10^-6)

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    2. For part b it is your answer for part a *(9.11*10^-31)

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    3. For part c it is you answer for part b /(1.602*10^-19)

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    4. I got d by using E=kQ/r^2 but it didn't really work out, but also it did haha
      multiply the answer for part c by the distance squared (0.0007)^2 and divide that answer by 8.99 (it should be by k=8.99*10^9 but it was off by 10^9 but I'm not sure why it didn't work exactly)

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    5. Hey, thanks for the walkthrough help! I haven't had a chance to complete this 'step by step' and your comment has helped a lot.
      I appreciate it

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    6. I think the reason that you have to multiply it by 10^9 is because the answer is in nC which is 10^9 smaller than C. So in order to be in the right units you have to divide the left side by 10^-9, effectively multiplying it by 10^9

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    7. where did you get 9.11*10^-31?

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    8. 9.11*10^-31 is the mass of an electron

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