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What is the change in the proton's electric potential energy?

  1. Apr 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A proton moves in a constant electric field E from point A to point B. the magnitude of the electric field is 6.4x10^4 N/C. The direction of electric field is opposite to the motion of the proton.

    If the distance from point A to point B is 0.50m, what is the change in the proton's electric potential energy, EPEb-EPEa?

    E=6.4x10^4 N/C
    Δd= 0.50
    ΔV=?

    2. Relevant equations
    ΔV=Δd*E
    V=W/q
    W=qEcosθ


    3. The attempt at a solution
    ΔV=Δd*E=(6.4x10^4)(0.50) = 32000J
    Clearly this is wrong, as the answer is supposed to be 5.1x10^-15J.
    Apparently you need to multiply it by the proton's charge, 1.602x10^-19C, and you would get the answer. but why? i dont see any formulas that would suggest me to multiply my answer by the fundamental charge.
     
  2. jcsd
  3. Apr 12, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    One of your equations is V=W/q. That means W=qV. For a proton q is not 1 coulomb. It's charge of a proton. And the units of Δd*E aren't joules. It's joules/coulomb.
     
  4. Apr 12, 2013 #3
    :) Arn't you fantastic!!! Thank you sir!!!
     
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