Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric Potential near a point charge.

  1. Sep 20, 2010 #1
    Hello,
    I am having trouble in proving the equation of the electric potential near a point charge Q. There are two different ways I can think to go about deriving the equation V=q/(4*Pi*E0*r). The first way is to calculate the change in potential energy of a test charge q0 in coming from infinite distance to a point P near the point charge Q, considering U=0 at infinity as a reference point. Then setting the change in potential energy equal to the negative of the work done by the electric field. The second way seems to me to be exactly the same mathematically. I am considering the change in potential energy in moving a test charge charge q0 from the point P to infinity. Again the change in potential energy should be equal to the opposite of the work done by the electric field along the path. Setting U=0 at infinity as a reference point, I find that the potential energy at P near the point charge Q is equal to the work done by the electric field along the path to infinity. Using the second method, I arrive at the right equation. But using the first method, I find a negative sign in the equation and I cannot get rid of it. I hope that somebody may be able to tell me what I have done wrong, or if there is something that I am not understanding correctly. I have attached a picture of my work on paper. I hope I have done a good job explaining the problem and that the picture is of decent enough quality.
    Thanks so much, hope to hear a response soon.
    -Alex
     

    Attached Files:

  2. jcsd
  3. Sep 21, 2010 #2
    The problem begins on the first line. The two equations cannot be the same. One time the scalar product is negative, and one time it is positive. The minus sign will flip the integral boundaries, but not the scalar product. I think the scalar product will do exactly what you are trying to force with the minus sign. We are running against the field so the result is automatically negative. No need for an extra minus.
     
  4. Sep 25, 2010 #3
    Yes you are right! thank you. I am still confused as to the cause of this however.
    Considering a positive point charge Q, the work done by the E field produced by Q along a radial line moving a positive point charge q0 from point a to point b, where a is closer to Q than b, should be the integral from a to b of F dot dr. Evaluating that integral I get a correct answer for the work, coming out positive. However if I calculate the work done by the E field produce by Q along the same line, but now from b to a, I would think the integral should be this: the integral from b to a of F dot dr... where now F is in the opposite direction as dr. But now, as you have said, switching the bounds creates a negative sign, and also the opposite direction of F and dr creates another negative sign... so the integral again comes out to be positive, but it is clear that is should come out negative! If you could point out the fault in my second integral I would be very very grateful, as this has been causing quite a lot of stress for me for the past few days, and I have an exam on monday so I would love to get this cleared up before then.
    I have attached a picture as well if it may be of some help to see what I am trying to describe.
    Thanks!
    Alex
     

    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook