Potential energy in concentric shells

AI Thread Summary
The discussion revolves around calculating the potential energy of an outer neutral shell due to an inner charged shell. It is established that the potential at the outer shell due to the inner shell is KQ/b, but since the outer shell has no charge, its potential energy is zero. Participants clarify that the electric field remains unchanged with the outer shell in place, and thus no energy is required to position it. The conversation also touches on the complexities introduced by the thickness of the outer shell and the concept of induced charges, emphasizing the need to reassess potential energy calculations in such scenarios. Overall, the consensus is that the outer shell's potential energy remains zero due to its neutral charge.
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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?
 
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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?
 
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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?
 
  • #31
Vouter=potential due to inner sphere+potential due to -Q(charge on the inner surface of outer shell)+potential due to +Q(charge on the outer surface of outer shell)

=##\frac{KQ}{3a}##+##\frac{-KQ}{a}##+##\frac{KQ}{3a}##
 
  • #32
why is the potential due to the charge on the inner surface of the shell -KQ/a on the outer surface?
 
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  • #33
Do you mean " a "is wrong?
 
  • #34
Yes, I mean that.
 
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  • #35
ehild said:
Yes, I mean that.
Distance between inner and outer surfaces of shell is "a "
 
  • #36
gracy said:
Distance between inner and outer surfaces of shell is "a "
What is the potential of the outer surface then if the shell is very thin? infinite?
 
  • #37
Thin or infinite?which one?
 
  • #38
gracy said:
Thin or infinite?which one?
Please, read my post.
 
  • #39
What is the potential of a sphere of radius R and charge Q at distance 0.0001R from its surface?
 
  • #40
ehild said:
0.0001R
0.0001 multiplied by R?
 
  • #41
Yes, at that distance. What is the potential?
 
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  • #42
ehild said:
What is the potential of a sphere of radius R and charge Q at distance 0.0001R from its surface?
@gracy ,
I added a few words to ehild's question.

This is asking,"What is the potential due to a (conducting) sphere of radius R and charge Q evaluated at distance 0.0001R from its surface?"(I may also post an additional reply to this thread approaching your difficulties from a different point of view.)
 
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  • #43
SammyS said:
This is asking,"What is the potential due to a (conducting) sphere of radius R and charge Q evaluated at distance 0.0001R from its surface?
R is not very big number right?
 
  • #44
SammyS said:
What is the potential due to a (conducting) sphere of radius R and charge Q evaluated at distance 0.0001R from its surface?
##\frac{KQ}{R}##
 
  • #45
gracy said:
R is not very big number right?
R may be any (positive) number.

0.0001R from the surface is a distance of 1.0001R from the sphere's center; barely outside of the sphere.
 
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  • #46
gracy said:
##\displaystyle \frac{KQ}{R}##
That's the potential at a location just inside the sphere, at a distance 0.9999R from sphere's center. Perhaps that is what ehild was asking for.
 
  • #47
You may want to add the following to the end of the OP.

"
The problem in this thread has been modified in Post #4 as follows:
gracy said:
Ok.There is a question.

A solid conducting sphere of radius, a, having a charge, Q, is surrounded by a neutral conducting shell of inner radius, 2a, and outer radius, 3a, as shown.Find the amount of heat produced when a switch connecting the spheres is closed.
"

gracy said:
I'm assuming that the region between r = 2a and r = 3a is the outer conducting shell.

The potential at the outer surface of the shell is given by ##\displaystyle \ V_\text{outer}=k\frac{Q}{3a} \ ##, assuming that potential → 0 as r → ∞ .

The potential at the inner surface of the shell is the same as ##\displaystyle \ V_\text{outer}\ ##, right ?

Find the potential difference at r = a and r = 2a, due to charge on the solid sphere.
 
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  • #48
SammyS said:
The potential at the outer surface of the shell is given by Vouter=kQ3a Vouter=kQ3a \displaystyle \ V_\text{outer}=k\frac{Q}{3a} \ , assuming that potential → 0 as r → ∞ .
What about induced charges ?@Tsny asked me to consider the induced charges as well
 
  • #49
Are you familiar with the potential V created by a total charge Q spread uniformly over a spherical surface of radius R? This is a fundamental result of electrostatics that every student should know. The potential created by this charge distribution has the following properties:

(i) For any point p outside the spherical surface, the potential is V = kQ/r where r is the distance from the center of the sphere to the point p.
(ii) All points inside the spherical surface are at the same potential V = kQ/R.
(iii) At points on the surface of the sphere, V = kQ/R.

It does not matter whether the spherical surface is a conductor or a nonconductor. It is assumed that we take V = 0 at infinity.

Use this along with the principle of superposition to find the potential at any point in your problem. You have 3 spherical charge distributions and the potential at any point is the sum of the potentials created at that point by each individual spherical charge distribution.
 

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  • #50
TSny said:
Use this along with the principle of superposition to find the potential at any point in your problem. You have 3 spherical charge distributions and the potential at any point is the sum of the potentials created by each individual spherical charge distribution.
That's what I have done in
ehild said:
why is the potential due to the charge on the inner surface of the shell -KQ/a on the outer surface?
I think it should be ##\frac{-KQ}{3a}##(using shell theorem)
 
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