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Electrostatics problem involving a Cone

  1. Mar 9, 2014 #1
    1. The problem statement, all variables and given/known data

    A cone made of insulating material has a total charge Q spread uniformly over its sloping surface. Calculate the energy required to take a test charge q from infinity to apex A of cone. The slant length is L.

    2. Relevant equations


    3. The attempt at a solution

    Potential at the tip of the cone is calculated by adding potentials due to rings of charges .

    Let us consider a ring having charge 'dq' at a distance 'x' from the tip and at slant height L'. L'=xsecθ where θ is the half angle of the cone .

    The surface area dS of the ring = 2π√(L'2-x2)dx

    dq=Q(dS)/(πRL)

    V = ∫dV = ∫kdq/L'

    This doesn't give me correct answer .I am not sure if I have approached the problem correctly.

    I would be grateful if somebody could help me with the problem.
     

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  2. jcsd
  3. Mar 10, 2014 #2

    TSny

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    Hello, Tanya. How "wide" is the ring on the surface of the cone? Is it dx?
     
  4. Mar 10, 2014 #3

    ehild

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    I can not follow your notations. What are x and L'?

    ehild
     
  5. Mar 10, 2014 #4
    Hi TSny

    The width is dxsecθ .That gives correct answer :smile: .

    But I am having difficulty in comprehending (rather visualizing ) the difference between using dx and dxsecθ :shy:

    My tiny brain can't differentiate between the two .I think if I use 'dx' then I am essentially covering the surface area of a cylinder ,not a cone . Am I right ?

    But how is surface area of ring given by (Perimeter)(slant height) ? What is the shape of the differential element ? Is the ring in the form of a cylinder or is it more like a frustum of a cone?

    'x' is distance of the center of the ring from the tip and L' is distance between the tip and a point on the circumference of the ring.
     
    Last edited: Mar 10, 2014
  6. Mar 10, 2014 #5

    BvU

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    You can help your brain by imagining a much bigger θ. Clearly no cylinder, hence the secθ.
     
  7. Mar 10, 2014 #6

    ehild

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    No need the centre of the ring mix in. The charge is distributed along the surface.
    You get a circular section when you unroll the side of the cone. The centre of the circle is the tip of the cone. Can you see how to get the distance of a surface point from the tip? Do you see how to get the area of a ring?




    ehild
     

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  8. Mar 10, 2014 #7
    Thanks ehild :smile:

    Please have a look at the attachment .I have edited your diagram .Have I represented the ring (in red) correctly ?
     

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  9. Mar 10, 2014 #8

    ehild

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    That "ring" is not a ring really but part of the cone surface. You do not need x at all. Let be l the distance of a point on the surface of the cone from the tip. The unrolled cone has the central angle Φ=2πr/L. Then the surface element is lΦ dl. The area of the side of the cone is A=L2Φ/2.

    ehild
     

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  10. Apr 3, 2014 #9
    Hello TSny...

    Could you please respond to post#4 and post#7 .

    I am still not very clear how dxsecθ is the width of the ring.I understand it is more of a calculus concept . But if you could give your views , that would be very helpful .
     
  11. Apr 3, 2014 #10

    TSny

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    Does the attached diagram help? If you "unwrapped" the strip off of the cone, it would be approximately a rectangle of width as shown in the diagram and length equal to the circumference of the circular cross section of the cone at the location of the strip.
     

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  12. Aug 16, 2017 at 2:02 PM #11
    It goes like this.
    Since u assumed an elementry part over the sloping surface , u must be inegrating over the surface .
    So,
    Instead of using dS= 2π√(L'2-x2) dx
    U must use dS=2π√(L'2-x2) dl.
    Hope u understood. This will give u correct answer.
     
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