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Homework Help: Potential energy and proton acceleratoin

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    A proton is accelerated to 3*10^6 m/s and heads towards a spherical charge distribution having a charge of 1.6*10^-18 C. Approximately how close to the distribution does the proton travel before stopping? (Assume inital potential energy of proton is zero).

    Vi= 0
    Q= 1.6*10^-18 C
    a= 3*10^6 m/s

    2. Relevant equations
    energy conservation:

    Kf+Vf = Ki + Vi
    V= kQ/R



    3. The attempt at a solution

    Kf+Vf = Ki + Vi
    0+qv= 1/2mv^2+ 0
    qv= .5mv^2
    (1.6*10^-18)(v)= (.5)(1.67*10^-27)(3*10^6m/s)
    v= .001565

    V= kQ/R
    .001565= (9*10^9)(1.6*10^-18)/(R)
    R= 9.2 * 10^-7m


    Is this correct?
     
  2. jcsd
  3. Apr 6, 2010 #2
    In EPE = qV, the 'q' is the charge being acted on by the spherical charge distribution, since you set V to be the potential caused by this distribution. So in this case it would be the charge of the proton.
     
  4. Apr 7, 2010 #3
    So when I plugged in 1 *10^-19 C [charge of proton] I ended up with 9.2*10^-6 m which was not even an option on the multiple choice. I am confused, please help. If this helps the answers are:
    1.9* 10^6m
    3.1*10^-13m
    5.5*10^-7 m
    9.2 *10^-7 m
     
  5. Apr 7, 2010 #4
    The charge of the proton is really 1.6*10^-19 C. But it looks like you made a mistake somewhere and lost a factor of 10. You actually had the right answer initially, but did the math wrong.

    So please write out all the steps and I can point out where you went wrong.
     
  6. Apr 7, 2010 #5
    so here's what I did:

    Kf+Vf = Ki + Vi
    0+qv= 1/2mv^2+ 0
    qv= .5mv^2
    (1.6*10^-19)(v)= (.5)(1.67*10^-27)(3*10^6m/s)
    v= .01565

    V= kQ/R
    .01565= (9*10^9)(1.6*10^-19)/(R)
    R= 9.2* 10 ^-8 m

    but then I did this:
    leaving q = 1.6*10^-18 and changing Q= 1.6 *10^-19 [ proton charge] and then solving the same way to get a final answer of
    9.2 *10^-7m.

    so which way is correct?
    By the way thanks for helping.
     
  7. Apr 7, 2010 #6
    Changing the order of q and Q should not affect your answer. You are making a mistake when you calculate R in this step:

    .01565= (9*10^9)(1.6*10^-19)/(R)

    Originally you have q = 1.6*10-19 C, and here in this step you are also setting Q = 1.6*10^-19 C. You should have set Q = 1.6*10^-18 C.
     
  8. Apr 11, 2010 #7

    well i found the right answer and it was nowhere near what I got.

    by the way i did originally have Q = 1.6*10^-18 C but you told me to change it.
     
    Last edited: Apr 11, 2010
  9. Apr 11, 2010 #8
    Well, you can interchange q and Q. But only one of them can be 1.6*10^-19 C. Unfortunately, you set them both to that.
     
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