1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric field and potential confusion

  1. Mar 20, 2009 #1
    Hey guys i am somewhat confused about Electric Field and Potential.
    I knew some day ago that although electric potential at a point is zero,electric field is not necessarily zero at that point. For example in case of a dipole, potential at right angle to axis of dipole is zero because theta is zero but at the same time electric field is not zero. How can it be? If electric field is the rate of change of potential then how it is possible that although potential is zero but field is non zero. In fact Potential=del(electric field).
    so please help
  2. jcsd
  3. Mar 20, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Why can't something have a non-zero derivative while having a value of zero?

    Consider the line y = mx + b. Does it have a non-zero slope (derivative)? Does it pass through a point where y = 0?
  4. Mar 20, 2009 #3


    User Avatar
    Science Advisor
    Homework Helper

    First of all, welcome to PF.

    The rate of change doesn't say anything about the value. For simplicity, let's work in one dimension, where the potential V(x) is a function of x only, and E = - dV/dx.

    Consider the potential V(x) = x.
    At x = 0, the potential is zero. However, it is clearly rising there: if you move a little away from x = 0 there is a change, so the electric field ~ rate of change of V is non-zero.

    The converse also holds. For example, suppose V(x) = 1 - x^2.
    At x = 0, the potential is non-zero, because V(0) = 1. However, the instantaneous change in the potential is zero (V'(0) = 0).

    For me it helps to view the potential as some "physical" object on which particles move, much like a rollercoaster in a theme park. In the first case, the potential is like a downward slope: if you put a particle on x = 0 it will slide downwards along the slope and never stop. In the second case, if you put a particle precisely on the top of the potential at x = 0, it will stay there. However, if you give it a slight push in either direction, it will slide down again.

    This analogy, although not perfect, is very helpful in several ways. For example, it explains why you can add an arbitrary constant to the potential (if you shift the entire rollercoaster up or down, if you put it on top of a mountain) the particle will still have the same behaviour; if you let a particle slide, it will go to the nearest (local) minimum (and if it has enough energy to overcome possible obstacles, it will go a global minimum, if any).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook