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Electric Field in center of half spherical shell

  1. Oct 29, 2007 #1
    A half spherical shell as seen in the picture.
    [​IMG][/URL][/IMG]

    charged with surface density [tex]\sigma(\theta)=\sigma_{0}cos\theta[/tex]

    Need to find the Electric field in center of the axis.

    As I see the direction of the electric field is the Z axis (because of the symmetric)

    to find the Field I use:

    dE[tex]_{z}[/tex] = [tex]\frac{kdq}{R^2}[/tex]

    Finding q, q=2[tex]\pi R^{2}\sigma_{0}cos\theta[/tex]?
     
    Last edited: Oct 29, 2007
  2. jcsd
  3. Oct 29, 2007 #2

    learningphysics

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    shouldn't Ez be [tex]\frac{kdq}{R^2}cos(\theta)[/tex]

    dq = [tex]\sigma(\theta)dA = \sigma_{0}cos\theta dA[/tex]

    what is dA here...

    plug your dq into your Ez formula... integrate from theta = 0 to pi/2.
     
  4. Oct 30, 2007 #3
    Why from 0 to pi/2?
    Can you explain?
     
  5. Oct 30, 2007 #4

    learningphysics

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    your [tex]dA = 2\pi*Rsin\theta(Rd\theta)[/tex]

    My element area is a circle of circumference 2*pi*Rsin(theta) multiplied by ds = Rdtheta. (see what I'm doing... I'm taking a circle of radius Rsin(theta))

    to cover the half circle theta needs to go from 0 to pi/2.

    to cover the whole circle you'd go from theta = 0 to pi.
     
  6. Oct 30, 2007 #5
    why do you multiply 2*pi*Rsin(theta) by Rdtheta?
    We need area so we must find the integral from circumference...

    And what is the logic taking circle 2*pi*Rsin(theta) as element of area?
    Why can't i just use 4[tex]\pi r^{2}[/tex] as the area of the shell?
     
  7. Oct 30, 2007 #6

    learningphysics

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    because your charge density depends on theta. so you need the charge at a particular theta...
     
  8. Oct 30, 2007 #7
    Thank you
     
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