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Zero Electric Potential Concept and Visualization

  1. Nov 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi, I've been having problems visualizing and interpreting a situation where there is zero potential in a point, equidistant, between two opposite charges. What is the significance of this? Here's a sample problem:


    Consider two point charges. One has a charge of +1 μC and the other has a charge of –1 μC. Consider a point P on the line connecting the charges. The point is the same distance from each charge. Which one of the following statements is true concerning the electric field and the electric potential at P?

    2. Relevant equations
    Electric potential formula: V= ke Qsource/r

    3. The attempt at a solution
    I know that at the equidistant point of the two opposite charges, the electric potential cancels out. Since electric potential is defined as the work done per charge of a positive charged particle to move it from infinity to a point under the influence of a source charge; does the zero potential between the two charges mean that it takes zero work to bring a charge from infinity to that point?
     
  2. jcsd
  3. Nov 23, 2014 #2

    haruspex

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    One problem is that electric potential is only really meaningful as a relative matter. Conventionally, you can define zero potential as the potential 'at infinity'.
    You can think of it as a landscape. Equipotentials form contour lines. If you move along contour lines there's no work done. If you move between two points at the same height, but crossing a valley or ridge to do so, there is still no net work done because the work you gained on the descent you lost on the ascent.
    Given (only) two equal and opposite charges, all points equidistant from them must lie on an equipotential. Since that includes points at infinity, we can say this is the zero potential.
     
  4. Nov 23, 2014 #3
    Alright, then if I moved a charge from infinity and traveled a path not along the equidistant equipotential line but ended up at a point at the equidistant equipotential line then no work is done still?
     
  5. Nov 23, 2014 #4

    haruspex

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    Yes.
     
  6. Nov 23, 2014 #5
    Thanks for your help!
     
  7. Nov 23, 2014 #6

    ehild

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    The work is the same as the electric field is conservative. The work done between two points does not depend on the actual path.
     
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