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[extremely easy question] the sign (+v or -ve) of potential energy

  1. Jan 24, 2012 #1
    since no one answer i edit it again.


    we got the forumla,

    v1-v0= the negative integral 0f E dot with dr from 0 to 1.

    So assume we are moving a positive point charge from infinity against the electric field emits by a positive source charge to location p.
    since E=f/q, and f dot dr = fdrcos180, then, the answer would be a positive number. is this true?
     
    Last edited: Jan 24, 2012
  2. jcsd
  3. Jan 24, 2012 #2
    Like charges repel, you have to apply a force in the direction you move the charge.

    Work = F dot x , both F and x point in the same direction.
     
  4. Jan 24, 2012 #3

    so, in this class, it would be positive right?


    the reason why I was confused because I messed it up with the ideal of gravitational potential energy. the infinity is always define as zero and the as the object fall toward the ground, the potential energy would always be negative and increase in lxl, what ever the value inside the magnitude sign.

    just having a one more quick question, when we say potential energy, for instance, compare -2 with -12, which one is larger?
    -2 is larger than -12? or do I take the abs value of them?
     
  5. Jan 24, 2012 #4

    SammyS

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    -2 is larger than -12, especially in this context.
     
  6. Jan 24, 2012 #5

    Nugatory

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    Easy rule of thumb: if you have to push on the object (positive charged particle being pushed towards another positive charged particle; pushing against a spring; lifting something heavy off the surface of the earth) you are increasing the potential energy, making it more positive/less negative. If you let it snap back (let go of your positive charged particle so it goes flying back out to infinity; let go of the heavy object so it falls back to the surface of the earth; let the spring relax) you are reducing the potential energy, making it more negative/less positive.
     
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