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Homework Help: Derivation of the potential of a charge

  1. Feb 1, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I'm facing an extremely easy problem (well, I set it up myself) that I can't even solve!
    I want to calculate the electric potential of a charge q in function of d, the distance from it.

    I let 0 be the point where the charge is.
    [tex]\varphi (d) - \varphi (0) = - \int _0^d \frac{kq \vec r}{r^3} \cdot d \vec r=-k q \int _0^d \frac{dr}{r^2}[/tex] and I would divide by zero if I continue.
    So I realize that instead of the 0 in the lower integral limit, why should I put [tex]\infty[/tex]? After all, it's not like setting the potential energy as 0 at infinity. I'm dealing with the electrostatic potential and not the potential energy (although they differ by a multiplicative constant in the static case).
     
    Last edited: Feb 1, 2010
  2. jcsd
  3. Feb 1, 2010 #2

    ehild

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    The potential at position r is defined as the work done by the field when a positive unit charge moves from r to the point where the potential is zero. In case of a point charge, we choose zero potential at infinity, so you have to integral from d to infinity.

    [tex]
    \varphi (d) - \varphi (\infty)= - \int _d^{\infty} \frac{kq \vec r}{r^3} \cdot d \vec r=-k q \int _d^{\infty} \frac{dr}{r^2}=\frac{kq }{d}
    [/tex]

    ehild
     
  4. Feb 1, 2010 #3

    fluidistic

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    Thanks once again ehild, for the nice clarification/explanation.
     
  5. Feb 6, 2010 #4

    fluidistic

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    Last question I hope: if the potential is positive, does that mean I have to do work to "push" the charge from r to infinity? And if the potential is negative, I don't have to do any work, rather the charge will do work and will get farer and farer?
     
  6. Feb 7, 2010 #5

    ehild

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    I copy my previous sentence :

    "The potential at position r is defined as the work done by the field when a positive unit charge moves from r to the point where the potential is zero. In case of a point charge, we choose zero potential at infinity"

    If the potential is positive (like in the case around a positive point charge U=kQ/r) the work of the field is positive when the positive unit charge moves toward infinity. Around a negative point charge, the potential is negative (because of the negative value of the charge), so the field does negative work if a positive unit charge moves to infinity, and this work decreases its kinetic energy. The KE never is negative. The positive charge can go away from a negative one if it has some initial KE or there is some other force balancing the electric field.

    When you move a charged particle in an electric field, there are two kinds of forces: The electric force and your force. The electric force is well-defined, you can find out that it is conservative, it has potential. Your force is as you like it (with limits, of course) you can not define the electric field by your force.
    Moreover: A particle does not do work on itself, it is the force that does the work.

    ehild
     
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