Investigating the Charge Distribution on a Bulging Sphere

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SUMMARY

The discussion centers on the charge distribution on a conducting sphere that develops a hemispherical bulge due to electrostatic forces. Participants explore whether the surface charge density (σ) remains constant and if the charge distribution over the bulge is uniform. It is concluded that while the initial charge distribution is uniform, the presence of the bulge complicates the distribution, suggesting it may not remain uniform due to the varying electric field and curvature effects. The conversation emphasizes the need for a simplified approach to understand the relationship between charge density and the geometry of the sphere.

PREREQUISITES
  • Understanding of electrostatics and charge distribution
  • Familiarity with surface charge density (σ) concepts
  • Basic knowledge of electric fields and potentials
  • Introduction to divergence in vector calculus
NEXT STEPS
  • Research the effects of curvature on surface charge density in conductors
  • Study the method of images for calculating charge distributions
  • Learn about the relationship between surface charge density and radius in conducting spheres
  • Explore the implications of electric field lines in non-uniform charge distributions
USEFUL FOR

Students and professionals in physics, particularly those focused on electrostatics, electrical engineering, and materials science, will benefit from this discussion. It is also relevant for anyone studying the behavior of charged conductors and their geometrical effects on charge distribution.

Daniel777
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Homework Statement
Due to nonuniform mechanical strength, when a conducting sphere is
given a charge Q , the electrostatic forces make a small hemispherical
bulge on its surface. The radius r of the bulge is negligibly small as
compared to the radius R of the conducting sphere.
Can you assume charge on the bulge distributed uniformly?
Relevant Equations
Charge distribution
In this question it is given that the sphere which is conducting is initially given a charge q then due to nonuniform mechanical strength and due to electrostatic force it creates a Small hemispherical bulge on its surface?

okay my doubt is Let me define a term σ where σ is surface density density so is dq/da going to be constant? And also on bulge part how would the charge distribution be is it going to be uniformly distributed over bulge part? According to me initially when sphere was given a charge q it was uniformly distributed but due to electric forces charges repel each other and create a bulge and I think they should repel each other and settle down in such a way that then also charge will be uniformly distributed only? Is it that?

I am really confused I can't be able to interpret the results? Please help me out?
 
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Hopefully you will get better replies from others, but it sure seems like if the growth in radius of the bulge is negligible compared to the sphere's starting radius, that the divergence would not deviate much in that hemisphere compared to the slightly smaller hemisphere. Maybe I'm missing something, though...

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https://mypages.iit.edu/~smile/guests/gsmxsec1.htm
 
Daniel777 said:
okay my doubt is Let me define a term σ where σ is surface density density so is dq/da going to be constant? And also on bulge part how would the charge distribution be is it going to be uniformly distributed over bulge part? According to me initially when sphere was given a charge q it was uniformly distributed but due to electric forces charges repel each other and create a bulge and I think they should repel each other and settle down in such a way that then also charge will be uniformly distributed only? Is it that?

I am really confused I can't be able to interpret the results? Please help me out?
I'm not sure whether you are saying the density should still be roughly the same over the whole sphere or just over the bulge (and different from that over the rest of the sphere).
Crudely, let's suppose the second case, ignoring variations over the bulge and local changes near the bulge.
For a given charge density, how does the surface potential vary with the radius of the sphere? For the whole surface of the sphere to be still at one potential, what does that suggest about the density on the bulge?
 
berkeman said:
it sure seems like if the growth in radius of the bulge is negligible compared to the sphere's starting radius, that the divergence would not deviate much in that hemisphere compared to the slightly smaller hemisphere.
Since r<<R, we can consider a ‘close-up’ view to get an intuition. The bump' and surrounding region is then, in effect, a conducting hemisphere attached to a flat conducting surface.

I interpret the question to be “Can you assume the charge on the hemisphere is distributed uniformly over the hemisphere?“.
 
Steve4Physics said:
I interpret the question to be “Can you assume the charge on the hemisphere is distributed uniformly over the hemisphere?“.
I was thinking more that it clearly cannot be (and likewise the distribution on the nearby plane/large sphere), but can it be so treated as a first approximation in order to get the relationship between the two densities ?
 
haruspex said:
I was thinking more that it clearly cannot be (and likewise the distribution on the nearby plane/large sphere),
I agree.

haruspex said:
but can it be so treated as a first approximation in order to get the relationship between the two densities ?
Possibly. interesting. I'm not sure though.

But the original question is:
"Can you assume charge on the bulge [is] distributed uniformly?"
and that seems a fairly specific/unambiguous question.
 
Steve4Physics said:
But the original question is:
"Can you assume charge on the bulge [is] distributed uniformly?"
and that seems a fairly specific/unambiguous question.
I feel that can be read as can we take it to be uniform for some (unstated) purpose?
 
haruspex said:
I feel that can be read as can we take it to be uniform for some (unstated) purpose?
Yes, that's a possible interpretation. I wonder if @Daniel777 (the OP) can confirm the question posted is accurate and complete. And if the official answer becomes available, it would be interesting to see it.
 
haruspex said:
over the bulge and local changes near the

Steve4Physics said:
Yes, that's a possible interpretation. I wonder if @Daniel777 (the OP) can confirm the question posted is accurate and complete. And if the official answer becomes available, it would be interesting to see it.
This is actually the original question.I hope it doesn't have any error.
 

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  • #10
Oh, it's a little bump, not the whole hemisphere of the overall sphere. Sorry that I misunderstood the question.
 
  • #11
berkeman said:
Oh, it's a little bump, not the whole hemisphere of the overall sphere. Sorry that I misunderstood the question.
Yeah but I am still in determining how would the charge configuration be over the entire sphere including the bulge part and electric field lines direction
 
  • #12
Is the divergence of D higher or lower above the bump compared to the rest of the sphere?
 
  • #13
berkeman said:
Is the divergence of D higher or lower above the bump compared to the rest of the sphere?
I haven't studied about the divergence yet
 
  • #14
Daniel777 said:
Yeah but I am still in determining how would the charge configuration be over the entire sphere including the bulge part and electric field lines direction
Typically charge distribution is a complicated matter except in some special cases. I feel confident saying It would not be uniform on either curved surface, but to solve the next part I think you have to assume it is near enough uniform. At least try that.
 
  • #16
PhDeezNutz said:
https://indjst.org/articles/capacitance-of-two-overlapping-conducting-spheres

Here’s a free paper on applying the method of images to overlapping spheres.
I couldn’t see how the method of images could be applied to hemispheres, though. Do you see a way?
Besides, I am only aware of such methods being used to find induced potentials. Finding a charge distribution is another matter.
 
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  • #17
haruspex said:
I couldn’t see how the method of images could be applied to hemispheres, though. Do you see a way?
Besides, Inam only aware of such methods being used to find induced potentials. Finding a charge distribution is another matter.
I may have just instinctively and prematurely drawn a comparison where there is none. I’m going to work on it…..maybe I’ll come up with something useful or maybe it will turn out that I was incredibly misguided.

Anyway thanks for challenging it because now I have an interesting problem to work on.
 
  • #18
Since @Daniel777 hasn’t met divergence yet, it sound like they are [choosing political correctness over grammar] working at an introductory level. That suggests only a relatively simple approach is being called for.

@Daniel777, have you met the effect of curvature on surface charge density (of conductors)? For two conducting spheres at the same potential, there is a simple relationship between the spheres’ surface charge densities and their radii.

(The classic example is that a sharp point on a charged conductor has a high surface charge density compared to the rest of the conductor’s surface.)

If you assume the average surface charge densities on the hemisphere and on the sphere obey this relationship, you might be able to answer part (b) of the question.
 
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  • #19
Steve4Physics said:
Since @Daniel777 hasn’t met divergence yet, it sound like they are [choosing political correctness over grammar] working at an introductory level. That suggests only a relatively simple approach is being called for.

@Daniel777, have you met the effect of curvature on surface charge density (of conductors)? For two conducting spheres at the same potential, there is a simple relationship between the spheres’ surface charge densities and their radii.

(The classic example is that a sharp point on a charged conductor has a high surface charge density compared to the rest of the conductor’s surface.)

If you assume the average surface charge densities on the hemisphere and on the sphere obey this relationship, you might be able to answer part (b) of the question.
This is what I have been trying to get @Daniel777 to try in posts #3 and #14.
 
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