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Calculating external work done in electric field

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data
    A region of space has a uniform electric field of strength 10.0 V/m due east. A small metal sphere with charge of +50.0 nC is in the field. There sphere is moved at a constant speed by an external force 10 cm north, 50 cm east, 20 cm south, and 50 cm east. How much work was done in total by the applied force?

    2. Relevant equations

    W = qEs

    3. The attempt at a solution

    W = qE, so I tried to plug in the charges and electric field information. But I don’t know how to apply the fact that the electric field is directed due east. I can calculate the work if it were moved AGAINST the electric field, but I don’t know how to calculate it if it were moved in the same direction as the electric field or perpendicular to it, as in this case.

    The total work is just the sum of all the individual work required to move the charge each time.

    Thanks for your help!
  2. jcsd
  3. Jan 31, 2009 #2
    If you have a force that's perpendicular to the direction of motion it does no work. It's the same for this. W=qEs, but F=qE, so W=Fs. If you apply the force perpendicular to the field, it does no work. So you only have to consider the east movements.
  4. Jan 31, 2009 #3

    The work is the same if I move it x amount east or x amount west, with the field due east?

    Part two of the question: What is its change in potential?


    Would the change in potential just be the amount of work done? Or is it zero since it is in the same electric field?
  5. Feb 1, 2009 #4
    Yes, the magnitude of work will at least be the same. I'm not 100% sure about it being positive or negative, but it's early in the morning and I can't think properly.

    No, Work= negative change in potential energy. Potential energy is related to potential.
    Last edited: Feb 1, 2009
  6. Feb 1, 2009 #5


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    Homework Helper

    If you are moving in the direction of the force then you have (-) work.

    Like for instance gravity going down gives back -work as kinetic energy.

    With work being given by the relationship:

    W = q*ΔV

    and your field given as 10v/m, if you move in the direction of the field you are moving toward lower voltage at the rate of 10v/m. So if you move 50cm + 50 cm toward negative voltage, your voltage will be lower by what amount? Since the ΔV is negative then your work should be negative shouldn't it?
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