Potential/self energy of a charged spherical shell

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Discussion Overview

The discussion revolves around the potential energy, or self-energy, of a uniformly charged spherical shell. Participants explore various formulas and concepts related to the potential energy associated with the shell, as well as the implications of charge movement and energy extraction from the system.

Discussion Character

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

Main Points Raised

  • Some participants propose that the potential energy (PE) of a charged spherical shell can be expressed as PE = QV/2.
  • Others present alternative formulas for potential energy, such as U = \frac{\sigma^2 \pi r^5}{9\epsilon_0} and U = \frac{Q^2}{8\pi \epsilon_0 r}, questioning the independence of potential energy from radius.
  • One participant discusses the conditions under which energy can be extracted from the shell, suggesting that whether energy can be given or taken depends on the ability of charges to move within the system.
  • There is a contention regarding the definition of "self-energy," with some stating it is the energy lost by charges if allowed to repel each other to infinity, while others emphasize the conventional nature of potential energy measurements.
  • Participants express concerns about the loose use of the term "energy," particularly in the context of gravitational potential energy of the universe.
  • A later reply provides a method for deriving the energy equation for a spherical shell, involving integrals over charge density and potential.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the correct expressions for potential energy and the implications of charge movement on energy extraction. There is no consensus on the definitions and interpretations of self-energy and potential energy in this context.

Contextual Notes

Some formulas presented depend on specific assumptions about charge distribution and the nature of the shell (conducting vs insulating). The discussion also highlights the ambiguity in defining energy levels and potential energy in a broader context.

Sourabh N
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What is the potential energy of a uniformly charged spherical shell; i think it is called the self energy of shell.
 
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The PE is QV/2.
 
How is the potential energy independent of radius?? I'm getting something like:

<br /> U = \frac{\sigma^2 \pi r^5}{9\epsilon_0}<br />

where, \sigma is the charge density of the shell.

OR:

<br /> U = \frac{Q^2}{8\pi \epsilon_0 r}<br />
 
Last edited:
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V=Q/4piepsilonzero r
 
How low can the potential go = how much energy we can extract from the sphere.

Can we squash it?

Can we stretch it?
 
Last edited:
Ulysees said:
How low can the potential go = how much energy we can extract from the sphere.

Can we squash it?

Can we stretch it?

??

The higher the potential energy, the more we can extract (if possible). If we squash it, we have to put in energy. By allowing it to stretch, we can get energy out of it.
 
Whether you can get energy out or give energy to the system depends on whether the charges can move.

If it is insulating and rigid so charges cannot move, you can get or give 0 energy.

If it is conducting but rigid, charges can only move inside the conductor. Therefore you can only GIVE some energy to the charges by forcing them to redistribute under external force.

So what is the "self-energy" of the shell then? By convention it's the energy that the charges lose if they are allowed to repel each other to infinity. Only by convention. Potential energy can be set to whatever level you want, it's only the changes that matter, what you get or what you give.
 
So the term "energy" is too often used loosely, we say things like "this object has this much energy" which is not strictly correct. Here's why.

Ill-stated question:

How much gravitational potential energy does the universe have?
 
rohanprabhu said:
How is the potential energy independent of radius?? I'm getting something like:

<br /> U = \frac{\sigma^2 \pi r^5}{9\epsilon_0}<br />

where, \sigma is the charge density of the shell.

OR:

<br /> U = \frac{Q^2}{8\pi \epsilon_0 r}<br />
The second formula is right, the first one is wrong. The question is about a sphere, not a ball, so one must insert Q=4*Pi*sigma*r*r, so
the energy is:

E=2*Pi*r^3*sigma^2/epsilon
 
  • #10
Ulysees said:
Whether you can get energy out or give energy to the system depends on whether the charges can move.

The total PE, calculated with the potential set to zero at infinity, can never be extracted, because we can never send all of the charge to infinity, but part of this total self-energy may be extracted. It's good that you have grasped the concept, because it was you who were asking the questions.
If it is insulating and rigid so charges cannot move, you can get or give 0 energy.
It can be broken up and the KE of the mutually repellent flying pieces may be collected to get some energy. Also, how did you get the charges in there in the first place? If that had required work to be done, perhaps the process can be reversed and the energy collected in a neater manner.

If it is conducting but rigid, charges can only move inside the conductor. Therefore you can only GIVE some energy to the charges by forcing them to redistribute under external force.

Just connect a wire to it and ground it, and collect part of the energy as the current flows.

So what is the "self-energy" of the shell then? By convention it's the energy that the charges lose if they are allowed to repel each other to infinity. Only by convention. Potential energy can be set to whatever level you want, it's only the changes that matter, what you get or what you give.
Who's saying anything to the contrary?

Ulysees said:
So the term "energy" is too often used loosely, we say things like "this object has this much energy" which is not strictly correct. Here's why.

Ill-stated question:

How much gravitational potential energy does the universe have?

I agree with you. But WHO exactly uses this meaningless question so often?
 
  • #11
Well, thanks for the answers and comments, but was expecting the method to derive the equation.
 
  • #12
> Just connect a wire to it and ground it, and collect part of the energy as the current flows.

You know as well as I know that current means the charge has to go somewhere. It can't go to infinity in all directions. So you can't ever extract the full energy that we have (by convention) associated with the shell.

>> Ill-stated question: how much gravitational potential energy does the universe have?

> I agree with you. But WHO exactly uses this meaningless question so often?

You must have missed the point. It's the bit in bold ("have") Not whether the whole question profoundly concerns the whole of humanity or something. "How much potential energy" is like saying "how close together can masses come theoretically".
 
  • #13
Sourabh N said:
Well, thanks for the answers and comments, but was expecting the method to derive the equation.

You never asked for the method...

The general formula for energy for a charge distribution is ½ [Integral over volume rho*V*(dxdydz)], where V is the potential at a point. To modify that for a spherical shell, we have to find W= ½ Integral over surface( s*V*da), where s is the surface charge density. The total charge is q and the radius is R., e = epsilon_nought.

V at the surface of the sphere = k*q/R, where k = 1/(4*pi*e). Then,

W = ½ kq/R Integral s*da = k/2(q^2/R) = [1/(8*pi*e)]*q^2/R.
 
  • #14
Ulysees said:
You know as well as I know that current means the charge has to go somewhere. It can't go to infinity in all directions. So you can't ever extract the full energy that we have (by convention) associated with the shell.

In post #10, for that very reason, I have written "part" in bold italics.

You must have missed the point. It's the bit in bold ("have") Not whether the whole question profoundly concerns the whole of humanity or something. "How much potential energy" is like saying "how close together can masses come theoretically".

Hey, I do get the point. I was just kidding a bit by asking WHO.

Cheers.
 

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