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Homework Help: A charged wire exerts a force on a proton.

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    An electrically charged wire on the z axis exerts a force on a proton, which moves on the x axis. The initial conditions for the proton position and velocity are:
    x(0) = x0 and v(0) = 0.

    The force on the proton is Fx(x) = C /x where C = 3.2×10−15 Nm.

    (A) Determine the potential energy function, U(x). Choose x0 to be the reference point; i.e., U(x0) = 0.
    (B) Calculate the velocity v at the time when the proton passes the point x = 2 x0.
    (C) Calculate the time t when the proton passes the point x = 2 x0. Assume x0 = 1 m.

    2. Relevant equations
    E= (1/2)m*v^2 + U(x)
    U(x)= -∫F(x)dx
    dx/dt= ±√[(2/m)(E-U(x))]

    3. The attempt at a solution
    In part A I integrated F(x) and got U(x)= -C*ln(x)
    When I got to part B I tried to find the energy assuming that since U(x0)=0 then KE(v0)=max=E but KE(v0)=0 meaning that energy would have to be zero and the velocity is an imaginary number so I'm lost on what to do. If someone could just point me in the right direction I think I could figure it out from there.
  2. jcsd
  3. Nov 6, 2011 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF,

    I'm not sure how you get that the velocity must be an imaginary number.

    It's perfectly fine for the particle to start out with zero total energy. It just means that as the force does work on the particle, its kinetic energy increases, and its potential energy decreases by a corresponding amount. In other words, at any time after the start, the kinetic energy is positive, since it increases from 0, and the potential energy is negative, since it decreases from 0. This means that the kinetic and potential energies are always equal in magnitude, and of opposite sign (such that their sum is always zero).

    To solve part b, find the change in potential energy between position x0 and position 2x0. The negative of this is the change in kinetic energy.
  4. Nov 7, 2011 #3
    Thanks alot man! I went back and realized when plugging in for U in dx/dt= ±√[(2/m)(E-U(x))] I kept forgetting to account for the fact that U is negative at x0. Stupid sign mistake!
  5. Nov 8, 2011 #4
    Do you have any ideas on how to solve part C? I'm trying to solve it by integration but I don't know how to integrate 1/√lnx, I've checked wolframalpha and that only showed how to integrate it unsing an imaginary error function. Is there some other method that I could use?
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