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Potential difference and a proton

  1. May 19, 2007 #1
    1. The problem statement, all variables and given/known data
    a proton (mass=1.67x10^-27kg, charge 1.6x10^-19C) moves from point A to point B under the influence of an electrostatic force only. at point A the proton moves with a speed of 50km/s and at point b it moves with a speed on 80km/s. determine the potential difference vb-va.


    2. Relevant equations
    delta V=delta U/Q
    also, in a uniform field delta V=-Ed
    F=qE=ma



    3. The attempt at a solution
    i don't know if enough information is provided in the question. i tried to find the acceleration of the proton as its speed increased from 50km/s to 80km/s but the question does not say how long it took to happen or how far it travelled in that time so using a=(v-u)/t cannot be found.

    Similarly, i can't find a way of obtaining the electric field because all we are given is the charge of the proton and the magnitude of the force is also not specified and can't be foind from f=ma since i can't get a.

    am i missing something or overlooking any details in the question? any help would be great
     
  2. jcsd
  3. May 19, 2007 #2

    hage567

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    Homework Helper

    I would say the kinetic energy gained by the proton is the same as the change in potential energy between the two points, which you can relate to the potential difference. Try that.
     
  4. May 20, 2007 #3
    hey THANKS :approve: i can't believe i didn't see that before!

    taking that approach, i used delta V= delta U/q
    after finding delta U from delta KE = - delta U
    so 1/2mv^2 final - 1/2mv^2 initial equalled 3.2565x10^-18, so change in potential energy was -3.2565x10^-18.
    dividing that by the charge of a proton i got +20V.

    can anyone tell whether my steps have brought me to the correct answer?
     
  5. May 20, 2007 #4

    Astronuc

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    Staff: Mentor

    20V is correct, however one should show the units when working a problem, particularly when there are mixed systems (MKS/cgs) involved, which is not the case here.

    Kinetic energy in the MKS system has units of kg-m2/s2 = Nt-m = J, and 1 V = 1 J/C, where C = Coulomb.
     
  6. May 21, 2007 #5
    ok i will definitely keep that in mind for next time.
     
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