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Deriving the formula for electric potential?

  1. Aug 31, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm getting really confused when it comes to electric potential. The definition I am using is 'the work done per unit charge on a small positive test charge when it is moved from infinity away to a point'. My first problem is that I'm not sure whether 'the work done' refers to the work done by the field or the work done by an external force.

    2. Relevant equations



    3. The attempt at a solution
    My second problem is that when I am calculating it, I don;t know whether to put a minus sign in front and why/why not:

    [itex]V=\frac{1}{q}\int_{\infty}^{r}\frac{kQq}{x^{2}}dx[/itex]

    I don't think there should be a minus sign because there is no minus sign in the equation for force, but I'm not sure.

    By third problem is that I am not sure which way the integral limits go. Since the charge is being moved from infinity away to the point, I think I have them righht but, again, I'm not sure.

    When I work it out, I get

    [itex]kQ\left [ \frac{-1}{x} \right ]^{r}_{\infty}=-\frac{kQ}{r}[/itex]

    Which only makes sense if the equatio refers to the work done by the field because then if the point charge creating the field is negative, the field does positive work to bring the charge closer...
    But this goes against my intuition because I think the potential should then be negative because the charge is getting trapped in a potential well (n the same way that the gravitational potential becomes more negative as the mass moves closer in).

    Thanks in advance for any help! I'm really confused :confused:
     
  2. jcsd
  3. Aug 31, 2014 #2

    ehild

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    Homework Helper
    Gold Member

    It is "the work that need to be done on a unit positive charge when it is moved from infinity to the point in question."

    But I prefer the definition with the work of the electric field:
    The potential at a point P is equal to the work done by the electric field while a unit positive charge moves from P to the point where the potential is zero.

    The electric field around a point charge Q is kQ/r2 and points away from the charge if it is positive. If you bring in a test charge from infinity you have to exert force opposite to the force of the electric field.

    So your work is [tex]W=\frac{1}{q}\int_{\infty}^{r}-\frac{kQq}{x^{2}}dx[/tex],
    so
    [tex]V=kQ\left [ \frac{1}{x} \right ]^{r}_{\infty}=\frac{kQ}{r}[/tex]

    If you use the other definition, the sign in the integrand is positive, but the integration goes from r to infinity.

    ehild
     
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