Electric Fields

Yeah, I hate this problem. I've seen it before. Okay, right off the bat you can say something about X and Y, since it is symmetric in those directions, right? Inside the sphere it should be 0 E field in the X and Y directions *I think*. Outside you have to calculate it, though... :(

 

What do you mean sigma = [02] _____ nC/m2 ?
What is the line for and why is [02] in brackets?

 

Hi daveyman,

I think the general procedure is to pick a general point on the surface you're integrating over. That point (actually a small area [itex]da[/itex]) will have a charge [itex]dq[/itex]. It will have spherical coordinates [itex]\{r,\theta,\phi\}[/itex] since they tell you to use spherical coordinates. What is the electric field contribution [itex]dE[/itex] from that charge [itex]dq[/itex] at the point you're interested in, in terms of [itex]\{r,\theta,\phi\}[/itex]? (Remember to use symmetry if you can to simplify it.)

Once you have [itex]dE[/itex], you can integrate over the variable coordinates over the entire surface to find the total field.

 

Yes, you are integrating over the area of the hemisphere, so it will be a two dimensional integral. (But one of them is straightforward, so you really only have to do work for one of them.)

The area comes in from the element dq, because you are given the charge/area [itex]\sigma[/itex]. If this were cartesian coordinates, you might have:

[tex]
dq = \sigma da = \sigma dx\ dy
[/tex]

but using the spherical coordinate area element for constant r gives:

[tex]
dq=\sigma da = \sigma r^2 \sin\theta\ d\theta\ d\phi
[/tex]

 

The exact factor will depend on how you set up your coordinates; is theta=0 at the rim or at the bottom?

Once you have that then the z component of dF will either be (dF sin(theta)) or (dF cos(theta)). But you have to look at your angles to be sure of which one.