# Homework Help: Need help

1. May 9, 2009

### en.yokhai

hey guys

i'm new here...i got here with a question...

i want to calculate the electric field in the middle of half sphere with raduse R ,
so..i made a rings out of the sphere,ds,which in fact is (2PI*R^2)sin(teta)d(teta), and tried to integrate from 0 to PI/2.

i did get something...i'd like to know if i'm right or not...

thanks:)

2. May 9, 2009

### diazona

I'd like to know what you got...

3. May 10, 2009

### en.yokhai

$$\sigma$$\2$$\epsilon$$

sigma dividd by 2 epsilon

4. May 10, 2009

### FedEx

well thats got to be wrong.. The electric field at any point is dependent on the radius vector... You equation has no r in it.... And why go for spherical co ordinates if rectangular co ordinates make it more easier...

The analogy which you have shown here is perfect... But remember that you have to take shells and not rings... And then apply k(charge on the infinitesimal shell) divided by x square... where x is the radius of the shell

5. May 10, 2009

### diazona

I get something else... (not $$\sigma/2\epsilon_0$$)

Try explaining how you did it.

Also, FedEx made me think of something: is this a solid half-sphere or a half-spherical shell? I'd assumed the latter because $$\sigma$$ is usually surface charge density.

6. May 10, 2009

### Cyosis

Are you talking about a spherical shell or a solid hemisphere?

7. May 10, 2009

### en.yokhai

8. May 10, 2009

### Cyosis

You forgot to select the correct component. In what direction does the E-field point in the center?

9. May 10, 2009

### FedEx

well i finally come to the conclusion that spherical coordinates would remain simple... And taking rings would be much better

but the rings would be like

dq = (sigma)(2pi r sintheta)(r dtheta)

And now apply the formula of the electric field of a ring

10. May 10, 2009

### orthovector

ahhh, yes. the infamous half sphere. you can't quite use gauss's law due to the lack of familiar symmetry, and it seems like you must integrate.

tell me about the source charge.