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Calculationg the electric field of an insulating sperical shell using integration?

  1. Feb 4, 2012 #1
    Calculationg the electric field of an insulating sperical shell using integration???

    1. The problem statement, all variables and given/known data

    We are asked to calculate the electric field at the center of an insulating hemispherical shell with radius R and a uniform surface charge density using integration.



    2. Relevant equations
    Gauss's law
    E*A = Qencl/epsilon

    dE = k(dQ/r^2) where k=1/(4pi*epsilon)


    3. The attempt at a solution

    I assume we can't use gauss's law since it asks for integration so I tried treating it as a ring of charge using dE = k(dQ/r^2)

    dQ = lamda * dS
    dS = rd(theta)
    dQ = lamda * r * d(theta)

    dE = k((lamda * d(theta))/r)(cos(theta) i + sin (theta) j)

    and integrating from there but i dont believe it can be treated as a ring of charge.

    Any suggestions??
     
  2. jcsd
  3. Feb 4, 2012 #2

    SammyS

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    Re: Calculationg the electric field of an insulating sperical shell using integration

    Hello nosmas. Welcome to PF !

    There's not enough symmetry yo use Gauss's Law.

    You should start by defining θ.

    Shouldn't you be using σ (sigma) for surface charge density ? ... or are you given The total charge, Q, so that σ = Q/(2πR2) ?

    dS = r dθ can't be correct. dS has units of area, r dθ has units of length.

    If the hemisphere sits on the xy-plane and is centered at the origin, then the x & y components of the E field cancel, leaving only the z-component.

    I does make sense to break the hemisphere up into rings of charge.

    Assuming θ is the angle of elevation above the xy-plane, then the width of such a ring is Rdθ. The radius of the ring is R cos(θ) . The area of such a ring is the width times the circumference.

    That may get you started in the right direction.
     
  4. Feb 6, 2012 #3
    Re: Calculationg the electric field of an insulating sperical shell using integration

    I should be using σ for the surface charge density I am just not sure how to incorporate it.

    So the E field on the hemisphere is equal to the sum of infinetly many rings of charge?

    I understand that the width of the ring is Rdθ and the radius of the ring is R cos(θ)

    so the area of the ring would be 2∏R(Rdθ) but I dont understand where to use the area?

    Using the general equation dE = K*(dQ/r^2) how would i incorporate σ into dQ?

    Thanks
     
  5. Feb 6, 2012 #4

    SammyS

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    Re: Calculationg the electric field of an insulating sperical shell using integration

    You left cos(θ) out of your expression for the area of a ring. (I inserted it in the above quote.)

    Charge = σ × area → dQ = σ dA = σ (2∏R)cos(θ)(Rdθ)

    I don't know how you have the hemisphere oriented in your coordinate system, but ... As I said in my previous post: "If the hemisphere sits on the xy-plane and is centered at the origin, then the x & y components of the E field cancel, leaving only the z-component." In fact this is true for the electric field, dE, due to each ring. So, to find dE, use Coulomb's Law multiplied by sin(θ) .

    Integrate the result to find E.
     
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