Potential energy in concentric shells

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SUMMARY

The discussion centers on calculating the potential energy of an outer neutral shell due to an inner charged shell. The potential at the radius of the outer shell is given by the formula V = KQ/b, where K is Coulomb's constant, Q is the charge on the inner shell, and b is the radius of the outer shell. However, since the outer shell is neutral, its potential energy is zero, as potential energy is defined as the product of charge and potential. The conversation also explores the implications of induced charges and the potential differences across a conducting shell with significant thickness.

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  • Understanding of electrostatics principles, particularly potential energy calculations.
  • Familiarity with the concept of electric potential and its relation to charge distributions.
  • Knowledge of the shell theorem and its application to spherical charge distributions.
  • Basic grasp of conducting and non-conducting materials in electrostatics.
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  • Study the derivation of electric potential due to spherical charge distributions.
  • Learn about induced charges and their effects on potential energy in electrostatic systems.
  • Explore the implications of the shell theorem in more complex charge configurations.
  • Investigate the energy calculations for systems involving multiple conductive shells.
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gracy
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Homework Statement


There are two concentric shells of radii a and b respectively ,inner shell has charge Q and outer shell is neutral what will be potential energy of outer shell due to inner shell

Homework Equations


##V##=##\frac{KQ}{R}##

PE=Charge .V

The Attempt at a Solution


I know what will be potential of outer shell due to inner shell it will be ##\frac{KQ}{b}## but as outer shell is neutral it does not have any charge and hence potential energy of outer shell should be zero,because we know potential energy is equal to charge multiplied by potential. charge is zero on outer shell hence it's PE would also be zero.Right?
 
Last edited:
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gracy said:

Homework Statement


There are two concentric shells of radii a and b respectively ,inner shell has charge Q and outer shell is neutral what will be potential energy of outer shell due to inner shell

Homework Equations


##V##=##\frac{KQ}{R}##

PE=Charge .V

The Attempt at a Solution


I know what will be potential of outer shell due to inner shell it will be ##\frac{KQ}{b}## but as outer shell is neutral it does not have any charge and hence potential energy of outer shell should be zero,because we know potential energy is equal to charge multiplied by potential. charge is zero on outer shell hence it's PE would also be zero.Right?
Sounds good to me.
 
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Hello Gracy,

Right !
It looks to me as if you sense a contradiction somewhere ?
The electric field ( in 'all of space' ) with the outer shell in place is identical to the electric field without, so putting it in place shouldn't require energy.
I wouldn't say
gracy said:
potential of outer shell due to inner shell will be ##\frac{KQ}{b}##
but

potential at r = radius of outer shell due to inner shell will be ##\frac{KQ}{b}##

[edit] this time haru was faster !
 
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Ok.There is a question.

A solid conducting sphere of radius a having a charge Q is surrounded by a conducting shell of inner radius 2a and outer radius 3a as shown.Find the amount of heat produced when switch is closed.
Heat produced=Initial energy -final energy
calculation.png


calc.png


I have taken interaction energy to be zero on the basis of the answer of OP.And I have considered formula of self energy of conducting and and non conducting spheres to be same.
I know formula of self energy of conducting sphere is ##\frac{KQ^2}{2R}##

Heat produced=##\frac{KQ^2}{4a}## - ##\frac{KQ^2}{6a}##

= ##\frac{KQ^2}{12a}##

But it is wrong!
 
In your original question, the outer shell was assumed to have negligible thickness so that the charge on the inner surface of that shell was essentially at the same location as the charge on the outer surface of the shell.

But now the outer shell has a significant thickness. You will need to reassess the initial potential energy.
 
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TSny said:
the outer shell was assumed to have negligible thickness so that the charge on the inner surface of that shell was essentially at the same location as the charge on the outer surface of the shell.
But I wrote the outer shell is neutral.So no charge in inner or outer surface of the outer shell.
 
The net charge is zero on the outer shell (initially). But, the inner and outer surfaces each have charge.
 
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TSny said:
The net charge is zero on the outer shell (initially). But, the inner and outer surfaces each have charge.
Are you hinting at induced charges?
 
gracy said:
Are you hinting at induced charges?
Yes.
 
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  • #10
TSny said:
But now the outer shell has a significant thickness.
Why do you think that ?There is nothing such mentioned in the problem.
 
  • #11
gracy said:
Why do you think that ?There is nothing such mentioned in the problem.
According to the statement of the problem, the outer shell has an inner radius of 2a and an outer radius of 3a.
 
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  • #12
TSny said:
You will need to reassess the initial potential energy.
What kind of changes shall I make in my calculations now?I don't know what is self energy of shells having significant thickness and I also don't know how to calculate interaction energy in such situation.
 
  • #13
If you have a system of static charges where an amount of charge Q1 is located where the potential is V1, an amount of charge Q2 is located where the potential is V2, and an amount of charge Q3 is located where the potential is V3, then the total potential energy of the system is

U = (1/2) (Q1V1 + Q2V2 + Q3V3)

This is a generalization of the formula for the potential energy of a capacitor: U = (1/2)QV.

Try to apply this to your system where you have three charges: charge on the sphere, charge on the inner surface of the shell, and charge on the outer surface of the shell.

U = (1/2) (QsphereVsphere + QinnerVinner + QouterVouter)

You should find that the last two terms simplify nicely. The work will be in getting the first term.
 
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  • #14
TSny said:
U = (1/2) (QsphereVsphere + QinnerVinner + QouterVouter)
I solved it as follows
last.png
But it is not correct where I went wrong.
 
  • #15
Why are the potentials at the surfaces of the outer shell different ?
 
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  • #16
I calculated it. Just the sign is opposite.
 
  • #17
gracy said:
I calculated it. Just the sign is opposite.
You did not show your calculation. The outer shell is a conductor. What do you know about the distribution of potential on a conductor?
 
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  • #18
ehild said:
What do you know about the distribution of potential on a conductor?
That it is same everywhere inside the conductor.
 
  • #19
ehild said:
You did not show your calculation.
CA.png
 
  • #20
gracy said:
That it is same everywhere inside the conductor.
Is it different on the surfaces of the conductor?
 
Last edited:
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  • #21
ehild said:
Is it different on the surface of the conductor?
No.
 
  • #22
gracy said:
No.
Then your calculation is wrong if you got different potentials for the inner and outer surfaces of the same conductive shell.
 
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  • #23
Yes but what is wrong in my calculation that I want to know.
 
  • #24
You can not calculate the potential so.
 
  • #25
ehild said:
You can not calculate the potential so.
I did not understand .potential of what?
 
  • #26
ehild said:
You can not calculate the potential so.
The potential of what have you calculated?
 
  • #27
ehild said:
The potential of what have you calculated?
I have calculates potential of the inner sphere and potentials of inner and outer surfaces of the outer shell.
 
  • #28
Why is the potential of the outer surface of the outer shell -kQ/(3a)?
 
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  • #29
I have calculated (my post #19)could you please tell me what is wrong in there?
 
  • #30
Explain the terms in Vouter. What are they?
 

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