# Electric Field in center of half spherical shell

1. Oct 29, 2007

### yevi

A half spherical shell as seen in the picture.
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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$$\pi R^{2}\sigma_{0}cos\theta$$?

Last edited: Oct 29, 2007
2. Oct 29, 2007

### 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.

3. Oct 30, 2007

### yevi

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

4. Oct 30, 2007

### learningphysics

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.

5. Oct 30, 2007

### 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?

6. Oct 30, 2007

### learningphysics

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

7. Oct 30, 2007

### yevi

Thank you

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