Electric field & force due to charged insulating hemispherical shells

  • #1
Aurelius120
196
20
Homework Statement
Two non conducting hemispherical surfaces of uniform charge density ##\sigma## are placed on horizontal surface. Find maximum compression in each spring if both surfaces are touching each other initially
Relevant Equations
Force on charge in an electric field=##qE##
Spring Force=##kx##
1000017600.jpg

So I know I have to equate force on a hemispherical shell with spring force to get value of compression but I can't find the force on the hemispheres
Some places that do have the solution use the formula :
$$\text{Field of non-conducting hemispherical shell= } \frac{\sigma}{2\epsilon_○} $$
This is also the formula for infinite sheet.
Then the net charge on other sphere is ##\sigma \pi r^2##
Then Option-C is correct answer

However I have two problems:
1. I couldn't derive the formula despite trying. It also doesn't makes sense that the field is uniform. The shell is made up of rings and rings have non-uniform fields. How do I understand and derive this?
2. The net charge on other hemisphere should be ##\sigma×2\pi r^2## Or ##\sigma×3\pi r^2##?
 
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  • #2
Since option D is "none of these", consider the possibility that what you need to do is prove A, B and C wrong. If so, you might not need to find an actual formula.
Note, I am not claiming that is the solution, merely a possibility.
 
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  • #3
haruspex said:
Since option D is "none of these", consider the possibility that what you need to do is prove A, B and C wrong. If so, you might not need to find an actual formula.
Note, I am not claiming that is the solution, merely a possibility.
The book's answer is option C though and the solution on the web is as described in the 1st post by somehow using the formula of infinite sheet
 
  • #4
You can find the formula for the field by using Gauss's law and the spherical symmetry.
 
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  • #5
nasu said:
You can find the formula for the field by using Gauss's law and the spherical symmetry.
Which field? If you mean the field due to one hemisphere, where’s the spherical symmetry?
Maybe consider the force on a surface element due to the rest of the sphere. (This is a little tricky: think about a test charge just inside the sphere and one just outside.) Then interpret that as a pressure, as though it is a bubble. That would give a force on each hemisphere.
 
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  • #6
You don't find the field of one hemisphere but of a sphere. As long as the two pieces "almost touch" the field is of a sphere. For a hemisphere alone there is no simple formula, of course. I don't think the problem in the OP expects to find such a field. Each piece of charge experiences the field of a full spherical shell.
 
  • #7
Aurelius120 said:
Some places that do have the solution use the formula :
$$\text{Field of non-conducting hemispherical shell= } \frac{\sigma}{2\epsilon_○} $$
A hemispherical shell does not have a constant field, not even considering only the field near the outher surface. So, there is no menaingful way of talking about "the field of..." and giving a constant value.
It looks like it may be the result of taking the field of a spherical shell (with unform charge density ##\sigma##) which is ##\frac{\sigma}{\epsilon_0}## right on the outside boundary and divide it by 2. It may be interesting to see the source of this quotation.
 
  • #8
nasu said:
You don't find the field of one hemisphere but of a sphere. As long as the two pieces "almost touch" the field is of a sphere. For a hemisphere alone there is no simple formula, of course. I don't think the problem in the OP expects to find such a field. Each piece of charge experiences the field of a full spherical shell.
A hemisphere cannot repel itself. By your logic, an isolated hemisphere would be subject to its own field as though it were external.
I believe the approach I outlined in post #5 gives the book answer .
 
  • #9
haruspex said:
A hemisphere cannot repel itself. By your logic, an isolated hemisphere would be subject to its own field as though it were external.
I believe the approach I outlined in post #5 gives the book answer .
I did not say that a hemisphere can repel itself. It does not follow from what I wrote. I said "each piece of charge feels the field of a full spherical shell". Of course, not full but minus the infinitesimal piece considered. But the point was that the field is that of a spherical and not hemi-spherical shell. You cannot write a constant value and associate it with "the field of a hemispherical shell".
It does not matter that the sphere is cut in the middle. The effect is something like an outward "pressure", as you mentioned in post 5. Only that for the OP question, you need to integrate the force components along the horizontal direction. Of course, the effect of one hemisphere on itself will cancel out in the integral. And the result is the force of the other hemisphere.
But to try to find the field of just one hemisphere and integrate the force produced by this on the other hemispehere would be a very ugly job.

Considering the pressure model you don't need to integrate at all, just take the projection of the area perpendicular to the direction of the desired force and multiply by the "pressure". I suppose that this approach is intended, if it is a non-calculus context. To find the pressure you need to consider a full sphere and not a hemisphere.
 
  • #10
nasu said:
Considering the pressure model you don't need to integrate at all
Of course.
nasu said:
each piece of charge feels the field of a full spherical shell

Well, not exactly. That's the tricky part I referred to in post #5.
 
  • #11
nasu said:
It looks like it may be the result of taking the field of a spherical shell (with unform charge density ##\sigma##) which is ##\frac{\sigma}{\epsilon_0}## right on the outside boundary and divide it by 2. It may be interesting to see the source of this quotation.
Solution 1
Solution 2
Solution 3
 
  • #13
nasu said:
It looks like it may be the result of taking the field of a spherical shell (with unform charge density ##\sigma##) which is ##\frac{\sigma}{\epsilon_0}## right on the outside boundary and divide it by 2. It may be interesting to see the source of this quotation.
Take a spherical shell with a small circular hole.
Consider a point just inside the hole. Were the sphere complete, there would be no field, so the actual field will be that -(the field that would have been created by the missing disc) ##= 0 -(-\frac{\sigma}{2\epsilon_0})=\frac{\sigma}{2\epsilon_0}##.
Now consider a point just inside the hole. Were the sphere complete, the field would be ##\frac{\sigma}{\epsilon_0}##
so the actual field will be that -(the field that would have been created by the missing disc) ##=\frac{\sigma}{\epsilon_0} -(+\frac{\sigma}{2\epsilon_0})=\frac{\sigma}{2\epsilon_0}##.
 
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