Potential in case of concentric shells

[SammyS]

No ,I thought it is charge supplied by Earth to shell C.

Yes, it's supplied by "earth", but it is also induced by the other charges.

Alternatively: The −q charge is indeed induced, whether the shell, C, is grounded (Earthed) or not. The grounding simply allows the +q charge, which would be on the outside surface of shell C to "escape".

Added in Edit:
Gracy,
EHild also responded to this, almost simultaneously (I guess she then deleted her reply). It's late here, so I will go off to bed now. You will be in very good hands with her.

 

[gracy]

Yes, it's supplied by "earth", but it is also induced by the other charges.

induced charge -q is on inner surface of shell C.And charge -q which has been supplied by Earth is on outer surface of shell C to balance induced +q charge there ,Am I right?

 

[ehild]

induced charge -q is on inner surface of shell C.And charge -q which has been supplied by Earth is on outer surface of shell C to balance induced +q charge there ,Am I right?

Well, this means that zero charge is on the outer surface of C, and -q induced charge on the inner surface. So that -q is induced charge.
In case of zero charge you can say that it is the resultant of one million C positive charge and one million C negative charge, but it has no sense at all.

 

[gracy]

Well, this means that zero charge is on the outer surface of C, and -q induced charge on the inner surface. So that -q is induced charge.

Yes,Am I right in post#52?

 

[ehild]

No. There is no charge on the outer surface of C. It is neutral. Charge means excess charge. If you have equal amount of electrons and single-charged positive ions, the charge is zero.

 

[jambaugh]

Pardon me for stepping in late but I would make the following observations on this problem. The potential for a spherically symmetric charge distribution will be V=\frac{kQ}{r} for Q the charge within the sphere of radius r, but only for a particular choice of gauge. The more general formula is V=\frac{kQ}{r}+G for G a gauge constant. Usually G=0 so that the potential at r=infinity is zero but you can set it to any value since it is only potential differences that matter. In this problem, you do not want the potential to change within the conductors and you want the potential of the grounded conductor to be zero.

Working in an arbitrary gauge can allow you to solve the problem a bit more easily.

Clearly the E field inside A is 0. The potential inside (and at) A is subject to our choice of gauge.

If q is the charge on A, the potential between A and B is \frac{kq}{r} + G with \frac{kq}{a} + G = V_A and \frac{kq}{2a}+G = V_B.

Since B has no net charge the potential between B and C is again \frac{kq}{r} + G. [Otherwise we might have to paste together distinct gauge conditions, different constants so potential is not discontinuous across a shell of charge density.]

This must at C give us V_C =\frac{ kq}{3a} + G, (lets say just as we reach the interior surface of C).

Using V_C=0 at ground to fix our gauge you get G = -\frac{kq}{3a} and so V_B= \frac{kq}{2a}-\frac{kq}{3a} = \frac{kq}{6a}.

Note however that as we cross a shell of charge (inner or outer to one of the conductors) we will need to adjust the gauge. For example between inner and outer surface of B shell, there is zero charge interior to spherical shells, the potential there is however not 0/r +G. It is the same as at the inner surface and the outer surface of B. But in each of these three locations you have distinct interior charge.

 

[gracy]

There is no charge on the outer surface of C

That's what I wrote

And charge -q which has been supplied by Earth is on outer surface of shell C to balance induced +q charge there ,

That is outer surface of shell has zero charge.

 

[gracy]

) What is the potential at a distance 4a from the origin due to the charge -q ?

$V$=$\frac{Kq}{4a}$?

What is the potential at a distance 3a from the origin due to the charge -q ?

$V$=$\frac{Kq}{3a}$

 

[ehild]

Pardon me for stepping in late but I would make the following observations on this problem. The potential for a spherically symmetric charge distribution will be V=\frac{kQ}{r} for Q the charge within the sphere of radius r, but only for a particular choice of gauge.

You are right, but Gracy wants to solve the problem with the Superposition principle, that is to get the potential as the sum of a single charged shells, A of radius a and charge q, and the other single single shell C of radius 3a and charge -q.

 

[ehild]

$V$=$\frac{Kq}{4a}$?

$V$=$\frac{Kq}{3a}$

These are correct. And the potential due the shell C is constant inside it, so what is that constant potential?

 

[gracy]

Please answer my post #57

 

[gracy]

These are correct.

I think I forgot negative sign

 

[gracy]

what is that constant potential?

$\frac{-Kq}{3r}$

 

[ehild]

I think I forgot negative sign

Yes, I also forgot. So it is -Kq/(3r) on the surface of C, and also inside.

 

[ehild]

That is outer surface of shell has zero charge.

Yes, that is correct, shell C has -q charge, but it is on the inner surface.

 

[gracy]

Then what's wrong in #52?

 

[ehild]

Then what's wrong in #52?

You wrote

And charge -q which has been supplied by Earth is on outer surface of shell C to balance induced +q charge there

Neither +q induced charge nor -q charge are on the outer surface. There is no charge there.
"Charge" is attribute of bodies. As they have mass, they can have charge. But the added or removed charge has no identity. You can not say that q and -q charges are together somewhere. The charge is (+q) +(-q)=0.

 

[gracy]

(+q) +(-q)

That's what I wanted to say what you wrote in the form of equation.I meant -q (supplied by earth)balances /makes zero induced charge +q.

 

[ehild]

That's what I wanted to say what you wrote in the form of equation.I meant -q (supplied by earth)balances (makes zero ) induced charge +q.

This is correct, but you can not say that those charges are on the surface.

 

[gracy]

but you can not say that those charges are on the surface.

yes,there is no charge anymore on the surface.

 

[gracy]

yes,there is no charge anymore on the surface

Then why we took this charge -q (supplied by earth) in calculation of net potential of sphere B if there is no net charge on outer surface of shell C?

 

[ehild]

Then why we took this charge -q (supplied by earth) in calculation of net potential of sphere B if there is no net charge on outer surface of shell C?

We did not. But there was -q charge on the shell C (on the inner surface) and we calculated with the potential due to it.
Once more: charge has no identity. The charge of shell A pushed q charge into the Earth from shell C, making that shell negatively charged.

 

[gracy]

I think I have got it.There were two questions in my mind while creating this thread
1)net potential of shell B
2)potential of shell B due to induced charge

To calculate these the most important thing we should know is charge distribution
Here charges in red are induced charge and charge in black was already present

This is just for clear understanding of charge distribution.But we know the induced charges on shell B cancel each other and -q (supplied by earth)balances /makes zero induced charge +q on shell C.That's why what we will actually see and the actual charges present are

Net potential is due to all charges present that is charge q shown in black on inner most shell (shell A)and charge -q show in red on inner surface of outermost shell (shellC)
Which we calculated as $\frac{Kq}{6a}$

And if question asks what is potential due to induced charge? we know -q shown in red on inner surface of shell C is induced charge hence potential due to that will come out to be $\frac{-kq}{3a}$
We should not get fooled by other red induced charge shown in first picture because those charges are already neutralized and no longer exist.

Am I right?

 

[gracy]

There is still one thing I don't understand that why don't the induced charges on shell c cancel each other as induced charges of shell B do?Why +q on outer surface of shell C is neutralized or balanced by charge coming from Earth and not by -q charge present on inner surface of shell C?

 

[ehild]

There is still one thing I don't understand that why don't the induced charges on shell c cancel each other as induced charges of shell B do?Why +q on outer surface of shell C is neutralized or balanced by charge coming from Earth and not by -q charge present on inner surface of shell C?

Charge has no identity.You can not say why that charge is balanced and not the other.
The -q charge of C (which is there as q charge was pushed into the ground by A ) is distributed as close to the positively charged A shell as possible, that is on the inner surface of C.

 

[gracy]

Thanks infinitely @ehild .God bless you all @SammyS @haruspex @gneill @cnh1995 @ehild @jambaugh

 

[gracy]

One last question

The -q charge of C (which is there as q charge was pushed into the ground by A ) is distributed as close to the positively charged A shell as possible, that is on the inner surface of C.

I just want to know why that -q charge stops on inner surface of shell C why it does not go beyond it.I mean if charge -q goes to inner surface of shell B it will be much more closer to +q charge.

 

[cnh1995]

One last question

I just want to know why that -q charge stops on inner surface of shell C why it does not go beyond it.I mean if charge -q goes to inner surface of shell B it will be much more closer to +q charge.

I believe the answer is in #12 and #13.There is already -q charge on the inner surface of B and +q on the outer surface, making E=0 in the thickness of B. So, +q responsible for emitting flux lines is actually on the outer surface of B. Hence,
-q on C "stops" as you say, on its inner surface.

 

[ehild]

One last question

I just want to know why that -q charge stops on inner surface of shell C why it does not go beyond it.I mean if charge -q goes to inner surface of shell B it will be much more closer to +q charge.

Charge is attribute, not entity. Only charged particles can move in the metal shell, (electrons) but they can not leave the metal. They stop at the surface.

 

[gneill]

One last question

I just want to know why that -q charge stops on inner surface of shell C why it does not go beyond it.I mean if charge -q goes to inner surface of shell B it will be much more closer to +q charge.

Charge needs a conductive path to follow in order to move unless the field is strong enough to overcome the work-function of the metal and "pull" the charge carriers off of it (electrons in this case), or if enough energy is supplied to the charge carriers for them to escape the surface. See, for example, the photoelectric effect, or thermionic emission.

As ehild has pointed out, In this problem we're considering charge as an attribute and not examining the mechanics of how that attribute is being shuffled around on carriers that have other properties as well (such as mass).

 

[gracy]

Ok.Got it!Thanks again @ehild @gneill @cnh1995

 

[gracy]

I find one more way to solve this problem.P;lease tell me which one is correct the one in #73 or this one or both?
To calculate these the most important thing we should know is charge distribution
Here charges in red are induced charge and charge in black was already present

But we know the induced charges on shell B cancel each other and the induced charges (+q and -q in red on shell C)will also cancel each other and then for the shell C to have zero potential Earth should supply charge -q .And hence there is net charge -q on shell C which has been supplied by earth.
.That's why what we will actually see and the actual charges present are

And the same thing as in post #73.But yes one more thing this charge supplied by Earth is also considered as induced charge.

 

[SammyS]

...

And the same thing as in post #73.But yes one more thing this charge supplied by Earth is also considered as induced charge.

It is the same thing as post #73.

If shells B & C were both neutral (no grounding of shell C), then the induced charge is the charge on each surface of those shells.

The grounding of shell C allows the charge on the exterior surface of shell C to be neutralized, so that the induced charge on the exterior of shell C is zero. In this sense, the charge "supplied" by the Earth is included in the induced charge.