Electric Field inside the material of a hollow conducting sphere

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Discussion Overview

The discussion revolves around the electric field inside a hollow conducting sphere when a positive point charge is placed within it. Participants explore the implications of adding or removing charges from the conductor and how this affects the electric field within the conductor's material. The conversation includes theoretical considerations, applications of Gauss's Law, and comparisons with insulating materials.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the electric field inside the conductor remains zero when additional positive charge is added to the outer surface after removing negative charge from the conductor's material.
  • Another participant asserts that the charges rearrange such that the electric field inside the conductor remains zero, regardless of the additional charge on the outer surface.
  • Concerns are raised about whether the extra positive charge creates a net electric field pointing inwards within the conducting material.
  • Gauss's Law is referenced, indicating that if a Gaussian surface is taken through the conductor's material and the extra positive charge is outside this surface, the electric field remains zero.
  • Participants discuss the implications of bringing an external charge close to a uniformly charged plastic shell and how it affects the electric field inside, noting that the symmetry of the charge distribution is crucial for applying Gauss's Law effectively.
  • One participant highlights that while the net flux remains unchanged, the electric field may vary in different regions due to the loss of symmetry when an external charge is introduced.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of the electric field inside the conductor when charges are added or removed. While some assert that the electric field remains zero, others question this assertion, indicating that the discussion remains unresolved regarding the implications of external charges on the electric field within the conductor.

Contextual Notes

Participants note that the application of Gauss's Law depends on the symmetry of the charge distribution, particularly in the case of insulators versus conductors. The discussion reflects uncertainty regarding the effects of external charges on the electric field in non-symmetric situations.

Who May Find This Useful

This discussion may be useful for students and enthusiasts of physics, particularly those interested in electrostatics, charge distribution, and the application of Gauss's Law in different contexts.

preachingpirate24
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Let's say I place a positive point charge inside a hollow conducting sphere. If we take a Gaussian surface through the material of the conductor, we know the field inside the material of the conductor is 0, which implies that there is a -ve charge on the inner wall to make the net enclosed charge 0 and a +ve charge on its outer wall. If I remove some electrons from the sphere, my textbook tells me that the +ve charge on the outer surface increases. Does this mean that the Electric Field inside the conductor is not equal to 0?

If it still is equal to 0, can someone please help me understand why? When I remove some negative charge from the conducting sphere's material, the positive charge on its outer surface becomes greater in magnitude, which to me implies that there is an Electric Field inside the conductor.

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I think I could also be misinterpreting the answer, and the author could be trying to say that the field inside the conductor was 0 before the extra charge was added to it, but then again, the author has stated that there can be no electric field inside a conductor in electrostatics. I understood why that statement is true, but when I add extra charges, I'm really not sure.

Thanks a lot for taking time to help :)
 
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If you put the charge inside, the charges of the conductor in the static state rearrange such there's no electric field inside the conductor, and there must be a surface charge distribution at the inner and the outer surface. From Gauss's Law you get that the inner surface must have a total charge of ##-4 \cdot 10^{-8} \text{C}##. Nothing changes on the inner surface of the conductor when putting the additional charge of ##6 \cdot 10^{-8} \text{C}## on the outer conductor but the additional charge distributes over the outer surface. Which thus must have a total charge of ##+10 \cdot 10^{-8} \text{C}##.
 
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vanhees71 said:
Nothing changes on the inner surface of the conductor when putting the additional charge of on the outer conductor but the additional charge distributes over the outer surface. Which thus must have a total charge of .

But wouldn't the extra positive charge create a net electric field pointing inwards in the conducting material? Or am I thinking along the wrong lines?
 
niranjanreji said:
But wouldn't the extra positive charge create a net electric field pointing inwards in the conducting material? Or am I thinking along the wrong lines?
Gauss's theorem says what is inside of enclosure matters, not outside. For example electric field is zero inside a sphere that has charged surface.
 
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niranjanreji said:
When I remove some negative charge from the conducting sphere's material, the positive charge on its outer surface becomes greater in magnitude
This is correct. So now apply Gauss’ law. What does Gauss’ law say will happen?
 
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Dale said:
This is correct. So now apply Gauss’ law. What does Gauss’ law say will happen?
If I take a Gaussian surface through the material of the conductor and the extra positive charge is outside the radius of this surface, the Electric Field is 0 since the net charge enclosed by it is 0. If I take a Gaussian surface with a radius larger than that of the larger sphere, I find that the flux is not 0, and hence the Electric Field is also not equal to zero. I'm pretty sure I'm right but I could be wrong here too.

I think I can see it better in my head now and justify it. Thanks a lot :smile:
 
niranjanreji said:
If I take a Gaussian surface through the material of the conductor and the extra positive charge is outside the radius of this surface, the Electric Field is 0 since the net charge enclosed by it is 0. If I take a Gaussian surface with a radius larger than that of the larger sphere, I find that the flux is not 0, and hence the Electric Field is also not equal to zero. I'm pretty sure I'm right but I could be wrong here too.
You are exactly right!
 
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Dale said:
You are exactly right!
I have a follow up question. What if I have a uniformly charged plastic shell with a positive point charge inside it at the center? The field is easy to find in this case using a Gaussian surface inside the plastic shell. But when I bring another positive charge close to the border of the shell, if I use the same Gaussian surface, the field inside doesn't change at all.

And this is even when the electrons in the plastic shell cannot rearrange themselves to cancel out the external field. Can I not apply Gauss's law when I'm working with an insulator?
 
preachingpirate24 said:
But when I bring another positive charge close to the border of the shell, if I use the same Gaussian surface, the field inside doesn't change at all.
You need to be careful here. Because the symmetry is disrupted only the net flux doesn’t change. The field will increase in some parts of the surface and decrease in others. The loss of symmetry prevents you from easily using Gauss’ law.

preachingpirate24 said:
Can I not apply Gauss's law when I'm working with an insulator?
It isn’t an issue about conductor vs insulator. The issue is the symmetry. With the symmetry Gauss’ law is easy, but without the symmetry you need to use numerical techniques.
 
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Dale said:
You need to be careful here. Because the symmetry is disrupted only the net flux doesn’t change. The field will increase in some parts of the surface and decrease in others. The loss of symmetry prevents you from easily using Gauss’ law.
What about the center of the plastic sphere then?
 
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preachingpirate24 said:
What about the center of the plastic sphere then?
Is the situation completely spherically symmetric?
 
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Dale said:
Is the situation completely spherically symmetric?
Yeah.. no it isn't. I suppose I'll learn how to solve this once I study some higher mathematics at college. Thanks!
 
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