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Find Change in the Electric Potential

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data


    A proton is released from rest in a uniform
    electric field of magnitude 40000V/m di-
    rected along the positive x axis. The proton
    undergoes a displacement of 0.3 m in the di-
    rection of the electric field

    The mass of a proton is 1.672623×10−27 kg

    A.Find the change in the electric potential
    if the proton moves from the point A to B.
    Answer in units of V.

    B.Find the change in potential energy of the
    proton for this displacement. Answer in units
    of J.


    C.Apply the principle of energy conservation to
    find the speed of the proton after it has moved
    0.5 m, starting from rest. Answer in units of
    m/s.
    2. Relevant equations

    V=Ed
    U=qV
    Kf-Ki+u=0
    k=.5mv^2
    Q

    3. The attempt at a solution
    A.) E=40000 V/m
    d=.3m
    mp=1.672623*10^-27kg
    q=1.60218*10^-19

    V=Ed
    V=(40000)(.3)=12000V ---but that is wrong

    B.)
    U=qV

    C.)
    Kf-Ki+U=0
    .5mv^2+U=0
     
  2. jcsd
  3. Sep 16, 2011 #2

    PeterO

    User Avatar
    Homework Helper

    (a) Electric potential is measured in volts. You were given V/m and metres. Just combine them to get volts [Joules per Coulomb; which you will need for part (b)]
     
  4. Sep 16, 2011 #3
    so what I started doing was right.... (40000V/m)(.3m) =12000 V

    when I put in that answer it says I am wrong :(
     
  5. Sep 16, 2011 #4
    Potential difference is given in volts. Electric fields are measured in V/m.

    E = dV/dx, so your change in potential dV = E*dx, which does match your working. However, since the electric field is doing work on the particle by pushing it, the particle is losing potential energy, not gaining it: your sign should be negative, I think.
     
  6. Sep 16, 2011 #5
    Thanks! I was just missing the negative sign.
     
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