Electric Field in center of half spherical shell

[yevi]

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$$\pi R^{2}\sigma_{0}cos\theta$$?

 

[learningphysics]

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.

 

[yevi]

Why from 0 to pi/2?
Can you explain?

 

[learningphysics]

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.

 

[yevi]

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?

 

[learningphysics]

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...

 

[yevi]

Thank you