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Electrostatics problem about part of a hollow sphere

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data
    A hollow sphere of radius R, centre at the origin, is cut into eight equal parts by 3 mutually perpendicular planes through the origin. Only one part is retained (the other 7 are removed) and given a charge q uniformly distributed on its surface. find magnitude of electrostatic intensity at the origin.

    Can u solve and explain how u got the ans plz?


    2. Relevant equations



    3. The attempt at a solution
    My teacher was looking at the resemblance of intensity of this part of the shell with that of a hemispherical shell, but I have no idea how that's possible.
     
  2. jcsd
  3. Sep 29, 2011 #2

    G01

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  4. Sep 30, 2011 #3
    But I have no idea how to approach this type of problem.

    I get that the charge would lie entirely on the curved surface and not on the plane surfaces, so net field would be due to that. In my opinion, the field should be same as that of a 8q charge on a complete sphere, but its 0 for a sphere, cause its a closed surface, and the part in question is an Open surface. i'm getting confused .
     
    Last edited: Sep 30, 2011
  5. Sep 30, 2011 #4

    G01

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    That may be but it still doesn't entitle you to a full solution and explanation of the problem. The reality is most people, when first confronted with a problem in any aspect of life, have no idea how to approach it. That what makes something "a problem." The ones who end solving said problem are those who don't let this stop them. That won't help you actually learn anything about problem solving, and I already know how to solve the problem, so there's no reason for me to go through typing up a full solution on this forum.


    Here's my hint: The field isn't zero anymore because the surface is not closed. There is no symmetry, so the field won't just be zero. There is a section in your physics book on finding the electric fields for continuous charge distributions. Find it. Re-read it. This is what the problem is about. How do they go about finding fields for continuous charge distributions? What methods do they use? Equations? Any calculus involved?
     
  6. Sep 30, 2011 #5
    I looked up and arrived at this:
    E of hemisphere with 4q = 4kq/2r² at center, upwards
    So, for 1/4 th of it, effective charge is q.
    shouldn't the E be kq/2r² now, in a direction radially outward?
     
  7. Sep 30, 2011 #6

    G01

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    OK, so that's a similar problem you can work from.


    Now, how did they arrive at that result? Could you apply the same method to this problem to see if you're correct?
     
  8. Sep 30, 2011 #7
    integration is always the last option, but in the solution sheet, it was derived from the results of hemispherical shell (vaguely explained)... seems like i'm having problem understanding the direction of E, but in my opinion, kq/2r² should be the resultant of all vectors?
     
  9. Sep 30, 2011 #8

    G01

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    You're magnitude for the field is incorrect. I think you're direction is correct, depending on what you mean by "radially outwards." Also, I think your result for the hemisphere with 4q is off by a factor of 1/2. I worked it out briefly, and it seems to me that it should be: [itex]E=\frac{4kq}{r^2}[/itex]

    Now, You have the results and derivation for the 1/2 shell sphere. This was almost certainly derived using some integration (no reason to be afraid of it!). This problem here is the same problem. The only thing that will change is the bounds of the integration and your total charge! What were they in that problem? What should we change them to?


    As for the direction of the field, you can find that from just symmetry considerations. If the 1/8 sphere, same as in the previous case of the 1/2 sphere. In the 1/2 sphere case, the field pointed directly away from the center of the hemispherical shell(which was along the z-direction). In this case, you should see that it behaves similarly, pointing away from the center of the shell.


    EDIT: Arnab, what level of physics are you in? I just want to make sure I tailor my help to your level from this post onwards.
     
    Last edited: Sep 30, 2011
  10. Oct 2, 2011 #9
    nope. google it. general expression for q on a hemisphere (shell) is [itex]E=\frac{kq}{2r^2}[/itex] i also derived it by considering rings as elements.

    im a 12th grader now, but such problems are generally not asked in school level, but only at competitive level.

    anyways, for figuring out E on 1/8 shell, i dont think it can be dont just by changing the bounds of integration. the only idea that comes to me is to consider quarter rings as elements. for that i think ill have to find E of quarter ring seperately?

    this is what i did for quarter ring, but i think is incorrect since im getting a pi^2 in the denominator. and integrating again with that for the shell will give higher powers of pi in the denominator...

    2mg8dvm.jpg
    for charge q on ring,
    σ=2q/∏r

    dE= ((kσr dθ)cos θ)/r^2 (im taking the cos component as the component in the direction of the resultant field).

    E=2∫<from 0 to ∏/4> ((kσr dθ)cos θ)/r^2
    =σ/(2∏εr) [sin θ] 0 to ∏/4
    = σ/(2√2∏εr)
    = q/(√2ε ∏^2 r^2)
     
  11. Oct 2, 2011 #10

    G01

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    Agh! Sorry, I dropped a factor of 1/2 when I was working it out on Friday. You are correct with that result. However, your result for the 1/8 shell is not correct.


    OK. Well, it can technically be solved by changing the bounds of a few integrals involved, but I'm certain I'm doing the problem in a way that is not appropriate for a high school level assignment.

    Let's start again and go about this in an entirely different (and much cooler, in my opinion) way:

    We won't need any calculus using this method. We know the field for the half shell with 4q is [itex]E=\frac{k4q}{2r^2}[/itex]. By symmetry, this field points in the z direction. Now, break that hemisphere into 4 corners, each with charge q. The charge is the same sign on each of the 4, 1/8th spheres. Thus, the z components of the individual corners are equal and add up to the total field for the hemisphere. Can you use this fact to find the z component of the field of one of the corners? I believe this is what you have already done above, but you only have the z component. You need the other two components!


    HINT: Now, if you think about it a bit, and consider that the field points directly away from the corner, you should be able to make an argument that the x- y and z components must all be equal.Putting these two pieces of information together, you should be able to then find the magnitude of the total field from the corner of the sphere.Try to see if you can get anywhere thinking about the problem along these lines. Let me know if you need any more help. I'm confident you'll be able to get the right answer this way, and have a great argument for it as well!
     
    Last edited: Oct 2, 2011
  12. Oct 3, 2011 #11
    Well, I get that charges will be now at corners, and the z component will be 1/4th of that of the hemisphere, and it will have x and y components also, since they are no longer canceled off by symmetry,

    But how to say that x & y components will be equal to the z component?

    Edit: ok I get it, but I need a deeper explanation
     
  13. Oct 3, 2011 #12

    G01

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    Do you agree that the symmetry of the problem dictates that the field must point directly away from the center of the corner shell? (i.e. normal to the center point on the shell)

    If the x,y, and z components were not all equal, then the field would not not point directly away from the corner. Carefully draw some diagrams if it helps you visualize it.
     
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