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- Thread starter daveyman
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What do you mean sigma = [02] _____ nC/m2 ?

What is the line for and why is [02] in brackets?

What is the line for and why is [02] in brackets?

- #4

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Poop-Loops - Yes, I'm sure you are right. The X and Y directions would be canceled out and therefore would be zero. So, the remaining field would exist only along the z direction. The question is what will be its magnitude?...

I'm pretty good with integrals, but I can't figure out how to set this one up. Basically, the integral needs to add up all of the vectors that point from the surface of the sphere to the origin.

Even a basic idea of how to set this up would help a lot!

Thanks!

- #5

alphysicist

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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.

- #6

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Great explanation! I think I'm getting the big picture now. One more question...

I've never done integration with spherical coordinates before, so I'm not quite sure how to handle it. The r stays constant, but both theta and phi will need to be variables - does this mean I need to integrate for multiple variables?

- #7

alphysicist

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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]

- #8

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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.)

How do I figure out how much of the field is in the z direction?

- #9

alphysicist

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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.

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