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

Homework Help: Potential Due to a Collection Charges

  1. May 16, 2008 #1
    My text reads for Potential Due to a Collection Charges:

    1/(4πє) ∑ q/r

    I'm consfused about the distance and how it's treated as a scalar.

    Let's say you want to calculated the potential between two protons. If a proton (or electron) is placed in the middle, there will be no net force and therefore no net electric field. With no electric field, you can have no potential, right? If you use the equation above, you get a potential.

    If the field point was 1 meter away from each proton, you would get the same potential as that of a field point 1 meter away from a particle with charge 2e(+).
  2. jcsd
  3. May 16, 2008 #2
    Ok, I think I figured it out. See if you can follow my logic.

    First of all, the potential is whatever you define it to be because you can add a constant.

    For the proton in the middle of two other protons, its potential is the same as a proton the same distance away from a particle of charge 2e+. However, this potential is a maximum for the case that the proton is between two protons and is constant everywhere between. For the latter case, the potential can increase if the field point is moved closer the the 2e+ particle.

    I used Gauss's Law to figure this out.
  4. May 16, 2008 #3
    That is not correct. With no electric field you have no potential difference. There can most certainly still be a potential.

    If you integrate the negative Electric field dot distance you get the potential difference. To get the potential at a certain point you just integrate from infinity to the point of interest.
    Last edited: May 16, 2008
  5. May 16, 2008 #4
    Right, initially I thought there was no field because I calculated it to equal zero. Then I realized that the potential is relative and can have whatever value. This material is best understood if you graph the potentials for the two cases.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook