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Nonuniformly charged spherical surface

  • #1
A sphere of radius a in free space is nonuniformly charged over its surface such that the charge density is given by ρs(θ) = ρs0 sin 2θ, where ρs0 is a constant and 0≤θ≤∏. Compute the total charge of the sphere.

So I know
ρs = dQ/dS

Integrating the surface charge density function will give me the charge Q. My question is how would you set up this integral?

∫ρs0 sin 2θ dS
integrating 0 to ∏

Or would this involve much more than that such as a triple integral?

Any help getting this set up would be great! Thanks!
 

Answers and Replies

  • #2
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This is not a triple integral. This is a surface integral.
 
  • #3
This is not a triple integral. This is a surface integral.
Alright then given the information above how would you set up the surface integral?
 
  • #4
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You have already done that. Now you just need to integrate it. You need to convert to to a double integral via an appropriate coordinate system. What would it be?
 
  • #5
You have already done that. Now you just need to integrate it. You need to convert to to a double integral via an appropriate coordinate system. What would it be?
I haven't had calculus in a long time so I have forgotten most of it.

∫ ∫ ρs0 sin 2θ dA = ps0 sin 2θ * 8∏r dr dθ
0 -> ∏ 0 ->a

A = 4∏r2
dA = 8∏r
 
  • #6
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It can't be true, it is a sphere, so r = a = const. What coordinate system do you think would be most appropriate here?
 
  • #7
It can't be true, it is a sphere, so r = a = const. What coordinate system do you think would be most appropriate here?
Well my first guess was a spherical coordinate system. Since it's a double integral it must be polar coordinates?

What would the limits then be?
 
  • #8
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You cant have "spherical" and "polar" at the same time.

The domain of integration is the entire sphere as follows from the problem.
 
  • #9
You cant have "spherical" and "polar" at the same time.

The domain of integration is the entire sphere as follows from the problem.
Sorry I think you misunderstood me because that wasn't worded clearly. I originally thought that spherical coordinates were going to be used (before I posted this problem, using triple integrals) but since you stated it can be done using a double integral my thought was to use polar coordinates. You are given the limits for θ. Typically in polar you would use r,θ? So then all I would need to do is find the limits for r?
 
  • #10
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You have a sphere. You can't describe a sphere in polar coordinates. You have a surface integral, where the surface is a sphere. You could use spherical coordinates, with which the surface integral over a sphere becomes a double integral. That's all there is to it, you just need to recall how you convert a surface integral to a double integral in a particular coordinate system. This is a very simple calculus problem, and I think you have studied the appropriate part of calculus.
 
  • #11
You have a sphere. You can't describe a sphere in polar coordinates. You have a surface integral, where the surface is a sphere. You could use spherical coordinates, with which the surface integral over a sphere becomes a double integral. That's all there is to it, you just need to recall how you convert a surface integral to a double integral in a particular coordinate system. This is a very simple calculus problem, and I think you have studied the appropriate part of calculus.
You are right and trying to pull it all back together is the hardest part. Does that involve parameterization described in terms of θ and Phi?
 
  • #12
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You are right and trying to pull it all back together is the hardest part. Does that involve parameterization described in terms of θ and Phi?
Since r is fixed (it's a sphere), then you have only the angles that are variable.
 
  • #13
Since r is fixed (it's a sphere), then you have only the angles that are variable.
So then Phi from 0 to 2∏ and θ from 0 to ∏? Is this what you are saying?
 
  • #14
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That's what you are saying, and that seems correct. But you need to express the surface element properly. That's part of the conversion from surface to double integral.
 
  • #15
That's what you are saying, and that seems correct. But you need to express the surface element properly. That's part of the conversion from surface to double integral.
Since r is constant then can I just say dS = r2 sin θ dθ dPhi?
 
  • #16
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This is correct.
 
  • #17
This is correct.
Alright that is good to know. So then should this be the integral to evaluate?

ρs0 * r2∫∫ sin(2θ) * sin(θ) dθ dPhi

0≤θ≤∏
0≤Phi≤2∏

After integrating that I get 2∏*ρs0*r2.
 
  • #18
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No, the does not seem correct. How do you integrate with respect to θ?
 
  • #19
No, the does not seem correct. How do you integrate with respect to θ?
For sin(2θ) * sin (θ) with respect to θ I got (2/3) sin(θ)^3.

Evaluated from 0 to ∏ I got 0.

Edit: TI89 got that answer which I have found isn't always reliable.
 
  • #20
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So if the integral over θ is zero, what is the end result?
 
  • #21
So if the integral over θ is zero, what is the end result?
Total charge = 0
 
  • #22
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Correct. Now, can you see that the density of charge in the northern hemisphere is exactly opposite to that in the southern hemisphere relative to the equator? The zero result follows from this antisymmetry.
 
  • #23
Correct. Now, can you see that the density of charge in the northern hemisphere is exactly opposite to that in the southern hemisphere relative to the equator? The zero result follows from this antisymmetry.
Yes that makes sense, they just cancel one another.

Thanks for the help!
 

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