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An electric dipole consists of a pair of equal but opposite charges

  1. Aug 10, 2014 #1
    1. The problem statement, all variables and given/known data

    An electric dipole consists of a pair of equal but opposite charges, +Q and -Q separated by a distance d. What is the electric potential at the point that's midway between these source charges?

    Through using the formula electric potential = kQ/r, I found the electric potential at P to be 0. Electric potential is the amount of potential energy each coulomb of charge would possess in an electric field right, so then why is it 0. If I place a negative charge at Point P, won't it be attracted to the +ve charge and repel by the negative charge. Won't it use potential energy to accelerate towards the +ve charge?
     
  2. jcsd
  3. Aug 10, 2014 #2

    haruspex

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    You're confusing potential with gradient of potential. The field is the gradient of the potential, and this is what causes the force on a test charge.

    I don't like the question, though. Potentials are not absolute. It's only potential difference that matters. You can take the potential at any point to be your reference potential (zero) and measure all other potentials in relation to it. The question only works if you specify also that the potential at infinity is zero.
     
  4. Aug 11, 2014 #3
    Does it matter that the point is 0 V only relative to other points as well. I don't think I am asking the question correctly, so I have attached a picture of the question which has the diagram too. It is part b).
     

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  5. Aug 11, 2014 #4

    haruspex

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    Not exactly sure what you are asking there.
    There is nothing special about 0V, it's an arbitrary reference point. As the attachment states, what matters is potential difference, i.e. the difference in voltage between two points.

    The attachment says the diagram shows "the electric potential at various points in the field they produce". That's a little misleading, because you could add some constant to every value and it would be effectively the same. On the other hand, it is customary to set the 0 V level as the potential "at infinity", which is perhaps assumed here. Or, equivalently, you could interpret the statement as meaning the charges add those potentials to an assumed background potential of zero everywhere.
     
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