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Sign on change in electric potential energy

  1. Apr 15, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm trying to figure out why potential energy decreases in this situation
    http://imgur.com/9PfjRVE [Broken]

    2. Relevant equations
    ΔU = FΔy

    3. The attempt at a solution
    If I define up to be positive, and ΔU = FΔy. In this picture, force is pointing down (so it is negative) and the height decreases so Δy is negative.
    Then ΔU = (-F)(-Δy) = FΔy. So then change in electric potential energy is positive, which means it should be increasing. But in the picture it says that U decreases.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 15, 2015 #2

    ehild

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    Force is negative gradient of the potential, so ΔU = - FΔy.
     
  4. Apr 15, 2015 #3
    Is there a reason it is negative? Everywhere in my book it keeps defining it as positive. And if it is anything like gravitational potential energy, shouldn't it positive? U = mgh for gravitational PE and U = qEy for electric potential energy.
     
  5. Apr 15, 2015 #4
    Think of this though: With the gravitational potential energy ##mgh## you don't use Final-Initial when you're calculating a work done. You do Initial-Final. Which is effectively ##-\Delta U ## If by ##\Delta## we understand a procedure of Final-Initial.

    For example: The work done by the gravitational field as an apple falls from a height of 5 m.

    ##m_{apple} (9.8m/s^2) (5m) - m_{apple} (9.8m/s^2) (0m)##
     
  6. Apr 15, 2015 #5
    Now that I think about it, Work = FΔy = -ΔU.

    So I'm assuming you can't set reference points anywhere you would like when it comes to electric potential energy? For example, if I made the reference point the height of an electron, and an electron moves below the reference point, I can't automatically assume that potential energy is negative?
     
  7. Apr 15, 2015 #6

    ehild

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    The potential energy in a point A is defined as the work done by the force when the body moves from A to the place where the potential is zero.

    So U(A)-U(0)=FΔr. U(A) is the initial potential energy and U(0) is the final one. We define the change as the final value - initial value. ΔU=U(final)-U(initial) so U(A)-U(0)= -ΔU: ΔU = - FΔr.
    In cases of gravity, U = mgh, and the force of the Earth points downward. In case of a falling body, the potential energy decreases.
     
  8. Apr 15, 2015 #7
    Adding to this, does the sign of the charge determine whether or not the potential energy is negative at this position?
     
  9. Apr 15, 2015 #8

    ehild

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    The electric field strength is the force exerted on a unit positive charge, and U is the potential energy of the unit positive charge. The force on a charge q is F=qE, and the work by the field when the charge moves from yA to yB is W=qE(yA-yB= U(A)-U(B)) = - q ΔU.
    The charge of the electron is negative. In the electric field, it moves in the direction of decreasing potential energy, which means increasing potential.
     
  10. Apr 15, 2015 #9

    ehild

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    The potential is given, but the potential energy of the charge depends on the sign of the charge .
     
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