Electric Field in center of half spherical shell

AI Thread Summary
The discussion focuses on calculating the electric field at the center of a half spherical shell with a surface charge density defined as σ(θ) = σ₀cosθ. The electric field direction is established as the Z-axis due to symmetry. The participants derive the electric field using the formula dEₓ = (kdq/R²) and discuss the integration limits from θ = 0 to π/2, which is necessary to cover the half shell. The area element dA is defined as 2πRsinθ(Rdθ), emphasizing the need to account for the variable charge density depending on θ. The conversation concludes with the understanding that the specific area calculation is essential for accurately determining the electric field.
yevi
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A half spherical shell as seen in the picture.
[URL=http://img219.imageshack.us/my.php?image=picnw3.jpg][PLAIN]http://img219.imageshack.us/img219/3424/picnw3.th.jpg[/URL][/PLAIN]

charged with surface density \sigma(\theta)=\sigma_{0}cos\theta

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_{z} = \frac{kdq}{R^2}

Finding q, q=2\pi R^{2}\sigma_{0}cos\theta?
 
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shouldn't Ez be \frac{kdq}{R^2}cos(\theta)

dq = \sigma(\theta)dA = \sigma_{0}cos\theta dA

what is dA here...

plug your dq into your Ez formula... integrate from theta = 0 to pi/2.
 
Why from 0 to pi/2?
Can you explain?
 
yevi said:
Why from 0 to pi/2?
Can you explain?

your dA = 2\pi*Rsin\theta(Rd\theta)

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.
 
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\pi r^{2} as the area of the shell?
 
yevi said:
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\pi r^{2} as the area of the shell?

because your charge density depends on theta. so you need the charge at a particular theta...
 
Thank you
 
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