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Homework Help: Changes in Energy

  1. Jan 26, 2015 #1
    1. The problem statement, all variables and given/known data
    (see picture)

    I understand one approach to this homework problem. I don't understand why an alternative approach fails.

    2. Relevant equations
    V = E*d

    3. The attempt at a solution
    The correct answer is (d), zero.

    An acceptable approach:
    1) Realize that V is the same at both corners
    2) Thus, delta V = 0
    3) delta U = q*V
    4) If potential energy doesn't change, then no work has been done (or been "lost") from the system.

    An alternative:

    1) The particle experiences a rightwards force in the direction of the 2+ charge
    2) To offset the right-wards pulling force and maintain a straight path, a new force, Fn, needs to be applied leftwards.
    3) The force (Fn) is applied over a distance (Not the vertical distance between both charges. Instead: The horizontal distance that would otherwise exist were Fn not exerted)
    4) This implies that energy has been exerted.

    Attached Files:

  2. jcsd
  3. Jan 26, 2015 #2


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    Staff: Mentor

    Sketch the force vector for different points along the trajectory from R to S. Work is done by the component of the force along the line of travel. Is there a symmetry involved?
  4. Jan 26, 2015 #3

    rude man

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    Homework Helper
    Gold Member

    Assume first the charge travels along the st. line R-S.
    Why would Fn be applied over a distance? Fn is applied just enough to keep the charge along R-S, which is perpendicular to Fn always.
    What is the vector formula for work?
    Finally, answer this: what if the charge moves along some path other than the st. line R-S. Now Fn would not be perpendicular to the path along much or most of the way, and the charge would move in the direction of Fn back & forth as the charge travels from R to S. So W would vary along this path. Why is it still true that net W=0, going from R to S?
  5. Jan 27, 2015 #4
    Doesn't the charge experience a net force pointing to the right?
    If it experiences a net force to the right, don't we have to exert a left-wards force to keep it moving down the center of the arrangement?

    The distance acted upon by the left-wards force isn't apparent, precisely because the force is inhibitive. It keeps the particle from moving the distance that we want to associate with energy.
    This logic obviously isn't sound. Why isn't it?
  6. Jan 27, 2015 #5


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    Staff: Mentor

    Force without displacement does no work. If the charge moves along a straight line perpendicular to the force that happens to be keeping it there, it does no work. This is akin to the way the floor provides a normal force to prevent an object from falling to the basement while you move it across the room. The floor does no work preventing the object from falling.

    What you need to be concerned about is the component of the net force that is directed along the line of motion of the charge. That part of the force will be doing work (positive or negative). Draw the sketch as I suggested.
  7. Jan 27, 2015 #6
    Ah this makes sense, especially with the example of the normal force exerted by the floor.

    Yes, there's a symmetry. I'll draw a picture soon so that future pf users have it for their reference.
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