Help Needed: Difficult Math Problem (3-4) Attached

  • Thread starter daveyman
  • Start date
In summary, this problem is hard because it requires integrating over a surface with charges distributed randomly. The student needs to understand how to set up integration with spherical coordinates and understand the electric field contribution for a point on the surface.
  • #1
daveyman
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This is probably one of the hardest homework problems I've ever had. Please see the attachment. The problem is 3-4.

This might be beyond the scope of this forum, but any help would be very appreciated!
 

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  • Homework 3.pdf
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  • #2
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... :(
 
  • #3
What do you mean sigma = [02] _____ nC/m2 ?
What is the line for and why is [02] in brackets?
 
  • #4
anirudh215, the sigma thing is just a variable. Each student gets a different set of input data so that everyone has slightly different answers. I usually just use the variable until the very end (sigma in this case) - then I insert the input data.

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
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.
 
  • #6
alphysicist,

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
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]
 
  • #8
Okay - one more thing...

alphysicist said:
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
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.
 

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The difficulty level of a math problem can vary depending on the person solving it. However, based on the given title, the math problem is categorized as difficult and falls within the range of 3-4 on a scale of 1-5, with 5 being the most difficult.

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