1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Challence problem on electric potential

  1. Apr 27, 2009 #1
    Edit to the title: CHALLENGE problem on electric potential energy*

    1. The problem statement, all variables and given/known data
    A proton and an alpha particle (q = +2e, m = 4u) are fired directly toward each other from far away, each with an initial speed of 0.010c. what is their distance of closes approach, as measured between their centers?


    2. Relevant equations
    u = proton mass = 1.67262158 × 10-27 kilograms
    e = elementary charge = 1.60217646 × 10-19 coulombs
    Kinetic energy Ke= 1/2 mv^2
    U = Kq1q2/r
    K = 9.0exp9
    deltaKe + deltaU = 0


    3. The attempt at a solution
    (deltaKe1 + deltaU1) + (deltaKe2 + deltaU2) = 0 where 1 represents the u mass proton and 2 represents the alpha particle.
    ((0-.5u(.01c)^2) + (Ke*2e/r)) + ((0-2u(0.01c)^2) + (Ke^2e/r)) = 0
    solve for r =2.45 *10^-14

    the answer in the textbook is 1.93 * 10 ^-14

    Iv done this over and over and over and am getting this 2.45 answer. please help!!!!! it is very frustrating! thanks!!!
     
  2. jcsd
  3. Apr 27, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi mrjoe2! :smile:
    ah … but they're not stationary at closest approach, are they? :wink:
     
  4. Apr 27, 2009 #3

    rl.bhat

    User Avatar
    Homework Helper

    If you consider a proton and an alpha particle as a system, at the closest approach the total kinetic energy of the system is converted to the potential energy.
    Hence
    .5u(.01c)^2) + + 2u(0.01c)^2) = (Ke*2e/r))
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Challence problem on electric potential
Loading...