Potential energy in concentric shells

[gracy]

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?

 

[haruspex]

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.

 

[BvU]

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

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 !

 

[gracy]

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

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!

 

[TSny]

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.

 

[gracy]

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.

 

[TSny]

The net charge is zero on the outer shell (initially). But, the inner and outer surfaces each have charge.

 

[gracy]

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?

 

[TSny]

Are you hinting at induced charges?

Yes.

 

[gracy]

But now the outer shell has a significant thickness.

Why do you think that ?There is nothing such mentioned in the problem.

 

[TSny]

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.

 

[gracy]

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.

 

[TSny]

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) (Q_sphereV_sphere + Q_innerV_inner + Q_outerV_outer)

You should find that the last two terms simplify nicely. The work will be in getting the first term.

 

[gracy]

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

I solved it as follows

But it is not correct where I went wrong.

 

[ehild]

Why are the potentials at the surfaces of the outer shell different ?

 

[gracy]

I calculated it. Just the sign is opposite.

 

[ehild]

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?

 

[gracy]

What do you know about the distribution of potential on a conductor?

That it is same everywhere inside the conductor.

 

[gracy]

You did not show your calculation.

 

[ehild]

That it is same everywhere inside the conductor.

Is it different on the surfaces of the conductor?

 

[gracy]

Is it different on the surface of the conductor?

No.

 

[ehild]

No.

Then your calculation is wrong if you got different potentials for the inner and outer surfaces of the same conductive shell.

 

[gracy]

Yes but what is wrong in my calculation that I want to know.

 

[ehild]

You can not calculate the potential so.

 

[gracy]

You can not calculate the potential so.

I did not understand .potential of what?

 

[ehild]

You can not calculate the potential so.

The potential of what have you calculated?

 

[gracy]

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.

 

[ehild]

Why is the potential of the outer surface of the outer shell -kQ/(3a)?

 

[gracy]

I have calculated (my post #19)could you please tell me what is wrong in there?

 

[ehild]

Explain the terms in Vouter. What are they?

 

[gracy]

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)

=$\frac{KQ}{3a}$+$\frac{-KQ}{a}$+$\frac{KQ}{3a}$

 

[ehild]

why is the potential due to the charge on the inner surface of the shell -KQ/a on the outer surface?

 

[gracy]

Do you mean " a "is wrong?

 

[ehild]

Yes, I mean that.

 

[gracy]

Yes, I mean that.

Distance between inner and outer surfaces of shell is "a "