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Electric Potential Problem Involving Uranium Atom

  1. Jun 12, 2013 #1
    Hi Everybody,

    I am currently battling with the fourth and final question of my A Level physics work for this week. The problem has been driving me crazy, as I seemed to have completed all of the relevant steps, and I find the right answer only with negative exponents, which isn’t much help at all.

    1. The problem statement, all variables and given/known data
    It then proceeds to ask questions for the electric force on the α-particles at the surface, and the electric potential of the nucleus, which I was fine with after a few tries. However, question (d) asks the following.

    2. Relevant equations

    [itex]W=\frac{kqQ}{r}[/itex] or [itex]E= \frac{V}{d}[/itex] together with [itex]V=\frac {kQ}{r}[/itex]


    3. The attempt at a solution

    I decided to tackle the problem in two ways, the first is as follows using the electric potential energy formula given above.

    [itex]W = \frac{qkQ}{r}[/itex]

    [itex]W = \frac{(8.99 * 10^9) * (3.2 *10^{-19}) * (1.44 * 10^{-17})}{(8 * 10^{15})}[/itex]

    [itex]W = \frac{(4.143 * 10^{-26})}{(8 * 10^{15})}[/itex]

    [itex]W=5.18 * 10^{-12}[/itex] J

    This is pretty much the same answer that I found in the back of the book, only with a negative exponent.

    My next attempt was a bit of a shot in the dark, but I tried it regardless.

    [itex]V = \frac{kQ}{r}[/itex]

    [itex]V = \frac{(8.99 * 10^9) * (3.2 * 10^{19})}{(8.0 * 10^{19})}[/itex]

    [itex]V = \frac{(2.877 * 10^{-19}}{(8 * 10^{-19}}[/itex]

    [itex]V = \frac{3.6 * 10^{-1}}V[/itex]

    [itex]E = \frac{V}{d}[/itex]

    [itex]E = \frac{(3.6 * 10^{-1})}{(7*10^{-19}}[/itex]

    [itex]E = 4.5 * 10^{18} J[/itex] J

    I would really appreciate it if someone could shed some light on this issue, as it has really been bugging me. Thank you very much for any help and your time.
     
    Last edited: Jun 12, 2013
  2. jcsd
  3. Jun 12, 2013 #2

    mfb

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    Staff: Mentor

    There are sign errors in your exponents.

    How do you get 8.0*1019 in the second approach?
    Some other exponents there look odd, too (how did you combine 9 and 19 to get -19?), and your units are missing or wrong.

    Just be more careful with your numbers and units, and all those mysteries will disappear.
     
  4. Jun 12, 2013 #3
    Thanks for the reply, mfb. The second solution was copied from the document that I was using for my work, and I had attempted that method numerous times, so I must have made a mistake with that approach. A fixed version can be seen below. The second solution seems to be completely wrong. I apologize for posting such incorrect workings earlier.


    [itex]V = \frac{kQ}{r}[/itex]

    [itex]V = \frac{(8.99 * 10^9) * (3.2 * 10^{-19})}{(8.0 * 10^{-15})}[/itex]

    [itex]V = \frac{(2.877 * 10^{-9}}{(8 * 10^{-15}}[/itex]

    [itex]V = {3.6 * 10^{5}}[/itex] V

    [itex]E = \frac{V}{d}[/itex]

    [itex]E = \frac{(3.6 * 10^{5})}{(8*10^{-15}}[/itex]

    [itex]E = 4.5 * 10^{19} [/itex]
     
    Last edited: Jun 12, 2013
  5. Jun 12, 2013 #4

    mfb

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    First, the formula E=V/d assumes that E is constant, and V is zero at a distance of d. Both are wrong.
    Second, why do you try to calculate the electric field here?
    Third, please add units to your calculations.
     
  6. Jun 12, 2013 #5
    In hindsight I shouldn't have included the second attempt , as even in my own mind it didn't seem right. What would the main issue be with the main attempt at a solution? Am I using the incorrect formula?
     
  7. Jun 12, 2013 #6

    mfb

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    Apart from the mentioned errors in the exponents, it looks right.
    Why do you think it would be wrong?
     
  8. Jun 13, 2013 #7
    It's strange because the back of the book has the answer as 5.18*1012. So I was able to get the right answer, only with a negative exponent.
     
  9. Jun 13, 2013 #8

    mfb

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    Probably a typo in the book.
    5.18*1012 J is roughly the energy you get in the explosion of 1000kg of TNT - certainly not the energy released in a single nuclear reaction!
     
  10. Jun 13, 2013 #9
    I thought that it seemed rather odd, but I thought that I had to be overlooking something. I did think that it was an awfully large number given the problem. Thank you very much for your help.
     
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