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 potential of a conical surface

  1. Mar 12, 2017 #1
    1. The problem statement, all variables and given/known data
    (From Griffiths problem 2.26) :
    A conical surface (an empty ice-cream cone) carries a uniform surface charge σ. The height of the cone is h, and the radius of the top is R. Find the potential difference between points a (the vertex) and b (the center of the top).

    2. Relevant equations

    3. The attempt at a solution
    Solution to this problem: (EXAMPLE 4 of the page)

    Whenever we are talking about electric potential,we need to specify the reference point.
    [itex] V(r) \equiv -\int_O^\mathbf{r} \mathbf{E \cdot} d \mathbf{l}\ [/itex]

    Where O is some standard reference point s.t. V = 0 at that point.In most circumstance we regard infinity as the reference point.

    If you picture an inverted cone in your mind, and follow the solution in the above link, then for V(0),
    you integrate from the base of cone to the tip of the cone.
    While for V(R) ( the base of the cone), you integrate from the tip of the cone up to the base of the cone).

    But this two directions are exactly opposite, so how can we have two different reference point when calculating the potential difference V(R) - V(0) ?
  2. jcsd
  3. Mar 12, 2017 #2


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    Not when I read it:
    For zv=0 they add all the contributions from the rings from z = 0 to z = R to find the potential V(0).
    For zv=R they add all the contributions from the rings from z = 0 to z = R to find the potential V(R).
  4. Mar 12, 2017 #3
    Yes, but since these two points are located at the limit of integration (z = 0 and z = R). For z = 0, the tip is at bottom,and all the contributions are above of the tip;
    But for z = R, all the contributions are below of the centre of base . I am a bit confused,does this invert the upper and lower limit of integration? And using the definition of V = [itex] V(r) \equiv -\int_O^\mathbf{r} \mathbf{E \cdot} d \mathbf{l}\ [/itex] , does the tip and centre of base have the same reference point ? I mean, because of "invert the upper and lower limit of integration" , I think the reference points in a and b are different as well. Sorry, I am very confused :nb)
  5. Mar 12, 2017 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No. The integral runs in the same direction in each case, from 0 to h.
    Yes. The reference point in both cases is zero potential at infinity. This is built into equation (16). Consider what happens to the expression for dV there as zv tends to infinity.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted