Potential in case of concentric shells

[cnh1995]

But that's we did .We did not involve these induced charges in calculation of final potential.

Yes, mathematically. But physically, they are involved in establishing that potential.

 

[haruspex]

But that's we did .We did not involve these induced charges in calculation of final potential.

I think the confusion stems from your reading of this para in post #17:

So, if you pick any point inside sphere B at a distance x from the center, at that point, the potential will be kq/2a. Out of that potential, kq/x will be contributed by +q on sphere A and the rest will be contributed by the induced charges on sphere B. ​

What cnh is saying there is that you can predict by other means what the net potential will be in the annulus, and know that the induced charges will be such as to achieve that. We do not ignore the induced charges, but we do not need to know what they are in order to find the potential.

 

[gracy]

Now what if question asks find the potential of shell B due to induced charges?will we able to calculate it?

 

[haruspex]

Now what if question asks find the potential of shell B due to induced charges?will we able to calculate it?

Yes. We know what the potential would be without induced charges, and we know what the net potential is with the induced charges. The difference is due to the induced charges.

 

[gracy]

Yes. We know what the potential would be without induced charges, and we know what the net potential is with the induced charges. The difference is due to the induced charges.

I thought we only know/calculated net potential of shell B i.e with the induced charge which is equal to $\frac{Kq}{6a}$

 

[haruspex]

I thought we only know/calculated net potential of shell B i.e with the induced charge which is equal to $\frac{Kq}{6a}$

That's what we have calculated, but we could also calculate the potential at B if there were no shell C.

 

[gracy]

You mean to know /calculate what the potential would be without induced charges we will have to ignore shell C but why?

 

[cnh1995]

If you are asked to calculate the contribution of the induced charges, simply follow what haruspex said in #34. There is no point in "neglecting" sphere C.

 

[gracy]

To follow what haruspex said in #34 we should know what the potential of B would be without induced charges.I am confused about how to calculate that.I mean that's what we did while calculating net potential of shell B because we did not involve induced charges in calculation of net potential of shell B.So we will get same result for potential of B without induced charges.And henece potential of shell B due to induced charges will come out to be zero

 

[SammyS]

To follow what haruspex said in #34 we should know what the potential of B would be without induced charges.I am confused about how to calculate that.I mean that's what we did while calculating net potential of shell B because we did not involve induced charges in calculation of net potential of shell B.So we will get same result for potential of B without induced charges.And hence potential of shell B due to induced charges will come out to be zero

I'm not clear as to exactly what you are intending to say here.

The potential (due to all charges) at a point on the surface of shell B is $\displaystyle \ k\,\frac{q}{6a} \$.

The potential (due only to the charge on shell A) at a point on the surface of shell B is $\displaystyle \ k\,\frac{q}{2a} \$.

The potential (due to all the induced charges) at a point on the surface of shell B is $\displaystyle \ -k\,\frac{q}{3a} \$.

 

[gracy]

why did you not involve charge -q which has been provided by Earth to shell C?

 

[haruspex]

why did you not involve charge -q which has been provided by Earth to shell C?

That was the third item, "potential due to all induced charges". The induced charges on shell B cancel each other.

 

[SammyS]

why did you not involve charge -q which has been provided by Earth to shell C?

I did.
That charge is at a distance of $\ 3a\$ from the origin. It produces no electric field at distances closer to the origin than $\ 3a\,,\$ and therefore the potential due to the that induced charge (-q on shell C), is constant at points closer to the origin than shell C.

Any effect from the induced charges closer to the origin than that (those on shell B) cancel each other out.

 

[gracy]

and therefore the potential due to the that induced charge (-q on shell C), is constant at points closer to the origin than shell C.

Then what is that constant potential?

 

[SammyS]

Then what is that constant potential?

I don't know why I didn't ask you first, but you should know that or know how to find that.

1.) What is the potential due to the charge -q at a distance $\ 4a \$ from the origin ( Common center) ?

2.) What is the potential due to the charge -q at a distance $\ 3a \$ from the origin ( Common center) ?

 

[gracy]

What is the potential due to the charge -q

Assuming charge q on shell C or you are asking in general?

 

[SammyS]

Assuming charge q on shell C or you are asking in general?

A charge of -q on shell C. That's the induced charge, right?

 

[SammyS]

I don't know why I didn't ask you first, but you should know that or know how to find that.

1.) What is the potential due to the charge -q at a distance $\ 4a \$ from the origin ( Common center) ?

2.) What is the potential due to the charge -q at a distance $\ 3a \$ from the origin ( Common center) ?

Maybe the wording is confusing?

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

2.) What is the potential at a distance $\ 3a \$ from the origin due to the charge -q ?

(The charge -q resides on shell C in both cases.)

 

[ehild]

Gracy, you should remember:
If charge Q is distributed uniformly on a spherical surface of radius R, its field and potential outside the sphere is the same as if the charge were concentrated in the center. Inside the sphere, the electric field is zero and the potential is constant.
The sphere in question is C now, the charge is -q and the radius is 3a. Outside the sphere, r≥3a, the potential due to this charge is k(-q)/r . On the sphere, it is k(-q)/(3a). If r<3a, the potential is constant. The potential is a continuous function, so it is k(-q)/(3a) everywhere inside the sphere, even on B, at r=2a.

 

[gracy]

A charge of -q on shell C. That's the induced charge, right?

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

 

[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?