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Griffiths (electrostatics) problem

  1. Jun 7, 2006 #1
    A conicalsurface(an empty ice cream cone) carries a uniform surface charge density sigma.The height & radius of the cone are h & R.Find potential difference between apex & centre of the top
  2. jcsd
  3. Jun 7, 2006 #2


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    Hi Kolahal and welcome to PF,

    Could please show your thoughts and what you have attempted thus far?

    HINT: Get your calculus books out. :wink:
    Last edited: Jun 7, 2006
  4. Jun 7, 2006 #3
    I have started the problem considering a ring ll to circular base.Expressed its radius in terms of its height,R,h.Then, I used the integral(Griffiths:2.30):-(1/4 pi epsilon not) integral over (sigma/curly r)da
    sigma= surface charge density,curly r=lr-r'l,da=area element
  5. Jun 7, 2006 #4
    OK, you mean this:
    [tex]V(r) = \frac{1}{4\pi \epsilon_0}\int \frac{\sigma}{r}da[/tex]
    Just a few hints:
    Express [itex]da[/itex] in terms of "r" and evaluate the electric potential at two points, viz. the apex and the centre of its surface.
    Last edited: Jun 7, 2006
  6. Jun 7, 2006 #5


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    So you have (just so I can see it more clearly);

    [tex]V = \frac{1}{4\pi\epsilon_{0}} \int \frac{\sigma}{|r-r'|} \; da[/tex]

    You need to express you change in area (dA) in terms of radii.

    Edit: Rashma got there before me.
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