# Calculate the electric field in the middle of half sphere

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:)

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diazona
Homework Helper
I'd like to know what you got...

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

sigma dividd by 2 epsilon

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

sigma dividd by 2 epsilon
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

diazona
Homework Helper
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.

Cyosis
Homework Helper
Are you talking about a spherical shell or a solid hemisphere?

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

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

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.