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Question about voltage in terms of dipoles

  1. Nov 8, 2008 #1
    If we have two charges, one at, say, 1C and one at, say, -1C, and they are separated by a distance of 1 meter, why is the voltage directly between them zero?

    A positive test charge will move towards the negative charge, and gain kinetic energy, but it started with no potential energy, and initial PE minus final KE has to equal zero, doesn't it? How can this be if you started with zero potential energy?

    I understand that voltage is potential, but you can multiply by the charge of the test point to get energy, can't you?
  2. jcsd
  3. Nov 9, 2008 #2


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    Staff Emeritus
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    Gold Member

    Voltage is potential difference, which is to say that it is the difference between the potential at two points, it is not defined for a single point - it can only be defined for a path.

    Perhaps you could clarify your question?

    Edit: I've just looked at the title of your post now and realised that your question relates to electric dipoles. Consider the diagram below
    The horizontal line represents an equipotential surface, in other words the potential of the electric field at any point along this line (or surface in 3D) has the same value. Note also that the potential is not uniquely defined, which means that there is now absolute potential, i.e. the potential must be measure relative to a path (or two points). This means that we can arbitrarily chose a point in space that we consider to have zero potential and measure the potential relative to this point.

    Do you follow?
    Last edited: Nov 9, 2008
  4. Nov 9, 2008 #3


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    Staff: Mentor

    No. (Initial PE + initial KE) equals (final PE + final KE). In your scenario, the positive test charge is initially halfway between the + and - charges. Its initial KE and initial PE are both zero, so the sum KE + PE is zero. Then it moves towards the - charge, under the influence of the electric forces exerted by the + and - charges. Its KE is now positive and its PE is now negative, and the sum KE + PE is still zero.
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