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Electric Feild Causing a Spark.

  1. Sep 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Approximately how large an electric field would be required to cause a spark in a gas at standard temperature and pressure if the ionization energy for the molecule of this gas is 1.5 times the ionization energy for air molecules, and the cross-sectional area of the molecules is 0.8 times the cross-sectional area of air molecules?

    2. Relevant equations
    eEd = U
    U = the amount of energy required to remove one outer electron from an oxygen or nitrogen molecule. (Air)
    e = 1.6E19 C
    d = mean free path

    3. The attempt at a solution
    I used the equation eEd = U, and tried to solve for E.

    I first determined U. In the book it gives U explicitly as 2.4E-18J. Since the question asks what happens when the ionization energy is 1.5 times the regular ionization energy for air molecules, i took (2.4E-18J)(1.5) = 3.45E-18 J

    Next, I found d. Since one mole of gas occupies 22.4 liters, that converts to 3.72E-26m^3 per mole of gas. I then used the equation d = (vol)/(pi*r^2). The approximate radius of one air molecule is said to be 1.5*10E-10m in the text. Therefore, since the question asks what happens when the cross-sectional area of the molecule is 0.8 times its usual area, I take (1.5*10E-10m)(0.8) = 1.2E-10m. After plugging in the volume and the new radius, d comes out to be 8.223E-7 m.

    Now I have all the numbers I need for: eEd = U
    E = (3.45E-18J)/[(1.6E-19C)(8.223E-7m).
    E = 2.62 E 7 N/C

    The problem is, that answer is wrong. The real answer is 3.6E6N/C (it is given in the back of the book). Please help, at this point I am unsure if I am even using the right equations. All I know for sure is that the equations HAVE TO have ionization energy and the cross-sectional area of a molecule in them, and you have to be able to solve for the E field. Thanks in advance for any help.
  2. jcsd
  3. Sep 9, 2008 #2
    Well I got the answer. All that was required was to multiply 1.5 and 0.8, and then take that times 3E6N/C. I really didn't know what I was doing, I just "reverse engineered" the answer in the book, and it turned out to be right (I had a similar problem on an online assignment, which is how I know that method works each time.) It seems so weird that a question so difficult sounding could turn out to be so easy. Why was nobody willing to help if the answer was so easy? So far I have posted two problems on here, only to have both go unanswered, and as a result I ended up losing points for one of them. How am I to get people to answer my questions, I give all the information asked?
  4. Sep 10, 2008 #3


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    As I said earlier in the PM, all our members offer homework assistance completely voluntarily and without recompense. We will help you with your homework, but are not here at your beck and call. The majority, if not all Homework questions are answered very quickly, but you may have to wait longer than eight hours for a reply. If your patient your questions will be answered.

    I looked at your other question, which you posted six hours before it was due in. Perhaps one shouldn't wait till the day one's homework is due in before asking for help?
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