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Potential of a charged cone

  1. May 14, 2015 #1
    How to calculate the potential at the apex of uniformly charged right circular cone (charge only at the curved surface), having height "h" and radius "R" and lateral height "l" and change density sigma?
  2. jcsd
  3. May 14, 2015 #2


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    The potential at a point [itex]\vec{r}_0[/itex] can be obtained by integrating [itex]\frac{dQ}{|\vec{r} - \vec{r}_0|}[/itex] over the charge distribution. If you have a surface charge distribution, then [itex]dQ = \sigma dA[/itex] where [itex]A[/itex] is the area, and [itex]\sigma[/itex] is the charge density.

    So the challenge is to find a convenient coordinate system for computing [itex]|\vec{r} - \vec{r}_0|[/itex] and [itex]dA[/itex] and then just do the integral. Do you have any ideas about a convenient coordinate system for a cylinder?
  4. May 14, 2015 #3
    Thanks for the hint... I tried it but its really getting quite difficult for me to choose a suitable area element for which i can calculate the potential and then integrating it over the whole area.... So can you please show me the steps to proceed... I need to get the question done today itself
  5. May 14, 2015 #4


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    Here's another hint: Suppose you split up the cone into narrow horizontal strips. Let [itex]r[/itex] be the distance of the strip from the apex. Let [itex]dr[/itex] be the width of the strip, and let [itex]L[/itex] be its length (the distance all the way around the strip). Then the area of the strip will be [itex]dA = dr \cdot L[/itex]. So can you figure out [itex]L[/itex] in terms of [itex]r[/itex]?
  6. May 14, 2015 #5
    Are you asking me to take the element as small strips wrapped around the cone as a circle?
  7. May 14, 2015 #6


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    I'm getting closer and closer to just giving you the answer. But here's another clue:

    If you have a cone that measures [itex]s[/itex] down the side, and the side makes an angle [itex]\theta[/itex] relative to the vertical, then the area of the cone's surface is:

    [itex]A = \pi s^2 sin(\theta)[/itex]

    So you can take the derivative with respect to [itex]s[/itex] to get [itex]dA[/itex]:

    [itex]dA = 2 \pi s sin(\theta) ds[/itex]

    That's the same as a little strip with width [itex]ds[/itex] and length [itex]2 \pi s sin(\theta)[/itex]

    So [itex]dQ = \sigma\ dA = \sigma\ 2 \pi s sin(\theta) ds[/itex]

    The distance from the strip to the apex is just [itex]s[/itex]. So

    [itex]\frac{dQ}{|\vec{r} - \vec{r}_0|} = \frac{dQ}{s}[/itex]

    So do the integral of [itex]\frac{dQ}{s}[/itex] where [itex]s[/itex] goes from 0 to [itex]l[/itex]. Can you figure out what [itex]sin(\theta)[/itex] must be?
  8. May 14, 2015 #7


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    @Quotes -- you need to start showing some effort, or this thread will be closed and deleted. We do not do students' work for them here at the PF.
  9. May 14, 2015 #8
    Sorry for this.. It was my first day at the forum. But yes for sure i will keep that in mind. Thankyou
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