Electric Field in center of half spherical shell

In summary, the conversation discusses finding the electric field in the center of a half spherical shell with a given surface density. The method used involves finding the element area and integrating from 0 to pi/2. This is necessary due to the charge density depending on theta.
  • #1
yevi
66
0
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 [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]?
 
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  • #2
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.
 
  • #3
Why from 0 to pi/2?
Can you explain?
 
  • #4
yevi said:
Why from 0 to pi/2?
Can you explain?

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.
 
  • #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?
 
  • #6
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[tex]\pi r^{2}[/tex] as the area of the shell?

because your charge density depends on theta. so you need the charge at a particular theta...
 
  • #7
Thank you
 

1. What is an electric field in the center of a half spherical shell?

The electric field in the center of a half spherical shell is the measure of the force per unit charge at the exact center of the shell. It is a vector quantity, meaning it has both magnitude and direction.

2. How is the electric field in the center of a half spherical shell calculated?

The electric field in the center of a half spherical shell can be calculated using the formula E = kQ/R^2, where E is the electric field, k is the Coulomb's constant, Q is the charge of the shell, and R is the radius of the shell.

3. What factors affect the electric field in the center of a half spherical shell?

The electric field in the center of a half spherical shell is affected by the magnitude and distribution of the charge on the shell, as well as the distance between the center of the shell and the point where the electric field is being measured.

4. Is the electric field in the center of a half spherical shell always zero?

No, the electric field in the center of a half spherical shell is not always zero. It depends on the distribution of charge on the shell and the distance from the center. If the charge is evenly distributed and the distance is large enough, the electric field will approach zero. However, if the charge is unevenly distributed or the distance is small, the electric field will not be zero.

5. What is the significance of the electric field in the center of a half spherical shell?

The electric field in the center of a half spherical shell is important in understanding the behavior of electric charges and the interaction between them. It can also be used in practical applications, such as in designing electrical circuits and devices.

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