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Electric potential equals the negative area under the graph

  1. Feb 23, 2012 #1
    When taking about potential and electric field, potential difference is equal to the negative of the area under the graph of E vs distance? why is that. My book defines it as the negative integral of Force times ds or V(intitial) - area under the curve. I don't understand why it's negative. I see it's the initial minus the entire are which would give me a negative potential difference, but why isn't it the final minus initial?
  2. jcsd
  3. Feb 23, 2012 #2


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    It's by definition, potential is defined to be the negative of the work done by a field. It makes sense when you consider situations where total energy = potential energy + kinetic energy, such as an object being accelerated by a field (with no losses). For example, a positively charged object being accelerated from the positive plate of a capacitor to the negative plate in a vacuum (no drag); as the object accelerates away from the positive plate, it's kinetic energy increases and it's potential energy decreases, and total energy remains constant.
  4. Feb 24, 2012 #3
    The negetive is result from:
    When a force take work,it transform the particular energy(such as electric inertial energy) into kinetic energy ,accoding to the conservation of energy, the particular energy is descreased.So f*s=-ΔE,so
  5. Feb 26, 2012 #4
    thanks everyone
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