Potential in case of concentric shells

[gracy]

Homework Statement

A,B and C are three concentric metal shells of radii a,2a and 3a .Shell A is the innermost and shell C is the outermost .Shell A is given a charge q and C is earthed.Find the final potential of shell B.

Homework Equations

##V##=##\frac{Kq}{r}##

The Attempt at a Solution

Charge q' is supplied from Earth to shell.

Then charges will be induced

Now we have to find q'
Potential of shell C should be zero.
##\frac{K(q'+q)}{3a}## - ##\frac{Kq}{3a}## +##\frac{Kq}{a}## - ##\frac{Kq}{a}## +##\frac{Kq}{2a}## =0

##\frac{K(q'+q)}{3a}## - ##\frac{Kq}{3a}## +##\frac{Kq}{2a}## =0

##\frac{K(q'+q)-Kq+\frac{3Kq}{2}}{3a}##=0

##K(q'+q)####-Kq##+##\frac{3Kq}{2}##=0

##\frac{2K(q'+q)-2Kq+3Kq}{2}##=0

##2K(q'+q)##-##2Kq##+##3Kq##=0

##2K(q'+q)##+##Kq##=0

##2K(q'+q)##=##-Kq##

##2(q'+q)##=##-q##

##q'+q##=##\frac{-q}{2}##

##q'##=##\frac{-q}{2}##-q

##q'##=##\frac{-3q}{2}##

Now by putting value of q' we can find potential of shell B by all the charges present.Am I right till here?

 

[cnh1995]

As per my understanding, you should consider "net charge" on the spheres. Potential on any sphere will be due to the net charge q on sphere A and q' on sphere C. In your first equation, your first term represents the potential of sphere C due to its own (net)charge, which is q'. Why have you taken q+q' then? Also, sphere B will have no net charge, hence will not contribute to any potential. So,why is there kq/2a in the first equation?

 

[theodoros.mihos]

Shells are thin so there is no radial charge distribution. There is only surface charge distribution that have the same value by spherical symmetry. Electric field between a and c can calculate be Gauss law and we know that potential on c is zero.

 

[cnh1995]

Sphere B is outside the sphere A but inside the sphere C. This means potential of B due to sphere C will be same as the potential of C due to its own (net)charge q'.

 

[gracy]

As per my understanding, you should consider "net charge" on the spheres. Potential on any sphere will be due to the net charge q on sphere A and q' on sphere C. I

Potential can also be due to induced charges.

 

[gracy]

Why have you taken q+q' then?

qi is supplied by Earth and +q is charge induced due to -q present on inner surface of shell C.

 

[cnh1995]

When charges are induced, both +ve and -ve charges appear on the sphere. When it is said that induced +ve charge flows to the ground, that means ground has supplied extra -ve charge to the body.

 

[cnh1995]

qi is supplied by Earth and +q is charge induced due to -q present on inner surface of shell C.

+q and -q will form net 0 charge. It will be only q' supplied by the ground that will be responsible for potential.

 

[haruspex]

##\frac{Kq}{a}##

Your expression for the potential at C is all wrong.
Hang onto these facts for uniformly charged conducting spherical shells:
- outside the shell, the potential and field generated by the charge on the shell is exactly the same as if the whole charge were concentrated at the centre of the sphere. (So what is the potential at C that results from the charge on A?)
- there is no field between the inner and outer radii of the shell, so the potential is constant in the annulus.
- there is no field generated by the charge on the shell inside the shell, so the potential it generates there is also constant and equal to the potential it generates at the shell itself. (Don't confuse this with the net potential at the shell.)

 

[gneill]

Gracy, if shell C were not grounded, do you agree that the charge induced on its outer surface by charge separation due to the charge on shell A would have been +q?

If so, then adding the ground connection would allow that +q charge to "escape" to ground. Or, thought of in another way, ground would supply a charge of -q to cancel that +q charge. The end result is that a charge of -q moves onto shell C through the ground path. This charge of -q serves to terminate the field lines of the interior charge +q of A shell, and so shell C exhibits a zero potential to the external world.

So you end up with shell A holding a net charge of +q, shell B holding a net charge of 0, and shell C holding a net charge of -q.

Given these shell charges you can apply the facts listed by @haruspex to find the potential at the location of shell B. Note in particular that an isolated uncharged shell (such as shell B) will not contribute to the potential. Only net charge on a shell will do so. The fields created by charge separation serve to cancel the field inside the conductor comprising the shell (so no currents continue to flow) and matches the external field at the boundaries of the conductor. That is to say, superposition of the external field and the field produced by the separated charges renders the net field inside the conductor (within the shell material) zero, but doesn't change the external field or potential.

In the end, the potential at the position of shell B will be the superposition of the potentials due to the net charges of shell A and C.

 

[gracy]

This charge of -q serves to terminate the field lines of the interior charge +q of A shell.

How?I mean electric field lines of interior charge +q of A shell have to be terminated on inner surface of shell B as electric field can not exist in metal body hence electric field lines of charge +q of A shell would not able to go beyond inner surface of shell B.

 

[cnh1995]

How?I mean electric field lines of interior charge +q of A shell have to be terminated on inner surface of shell B as electric field can not exist in metal body hence electric field lines of charge +q of A shell would not able to go beyond inner surface of shell B.

Electric field will not exist in the thickness(however small it is) of the sphere. So, when -q charge is induced on the inner surface of B, +q charge is induced on the outer surface. This makes the field 0 inside the thickness of the sphere. However, as now +q is on the outer surface of B, its field lines will reach out to sphere C. Here, there is a 'discontinuity' in the field due to the thickness of the sphere. The field wouldn't be trapped inside any sphere, if sphere C weren't grounded.

 

[haruspex]

How?I mean electric field lines of interior charge +q of A shell have to be terminated on inner surface of shell B as electric field can not exist in metal body hence electric field lines of charge +q of A shell would not able to go beyond inner surface of shell B.

Yes, you can think of it as lines from A terminating on inner surface of B, but then being regenerated from the outer surface of B. Or, as gneill suggests, you can note that B has no net charge, so (as far anything except B itself is concerned) it is the same as the lines from A going all the way out to C. Work with whichever you are comfortable with.

 

[gracy]

Note in particular that an isolated uncharged shell (such as shell B) will not contribute to the potential. Only net charge on a shell will do so.

Then why there is potential due to induced charges in the video I have mentioned?

 

[cnh1995]

The charge in the video is external to the sphere. Hence, to make the potential of every point inside the sphere same, there will be potential due to the induced charges. The net potential of every point inside the sphere will be kq/r. Out of that, kq/x will be due to the external +q and the rest will be due to the induced charges. In fact, to make the potential equal to kq/r, the charges will be induced.
In case of concentric spheres(your original question), the charge is inside the spheres B and C and 'on' the sphere A.
Consider only spheres A and B:
Charges will be induced on inner and outer surface of B to make the potential inside B equal at every point, thus making
E_{inside B}=0. 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. But
the net potential of sphere B is kq/2a. That is what is asked in your original question.
I haven't considered sphere C here. You can extend this logic to sphere C and get the final answer.

 

[cnh1995]

In short, potential of the sphere is decided by the net charge present in the system(and radius of the sphere(s)). Induced charges only help to establish this pre-decided potential, wherever necessary. At the center of the sphere(in the video above), contribution due to the induced charges is 0 since only +q charge outside is sufficient( or +q is the one to decide the potential at the center) and the induced charges are symmetrical and equidistant from the center. Anywhere else inside the sphere, potential due to the +q outside is not kq/r. So, to make it kq/r, charges are induced on the sphere. So, in your original question, it is only the net charge that will decide the potential of the spheres. So, net charge is the one who will decide the potential and induced charges will simply help to execute the task of establishing that potential everywhere inside the sphere and make E_inside=0.

 

[gracy]

The net potential of every point inside the sphere will be kq/r. Out of that, kq/x will be due to the external +q and the rest will be due to the induced charges.

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.

I am not getting then how potential only depends on net charge present and not on induced charges .Isn't induced charge playing role equal to that of net charge present in determining potential?

 

[cnh1995]

I am not getting then how potential only depends on net charge present and not on induced charges .Isn't induced charge playing role equal to that of net charge present in determining potential?

Magnitude of the net potential is decided by the net charge. Induced charges appear to establish that potential inside the sphere, wherever necessary. So, role of induced charges is vital in "establishing" the net potential(which is already determined by the net charge) and not in determining it. Of course, the potential due to induced charges will vary at each point but the total potential at any point will remain same i.e. kq/r.

 

[haruspex]

Consider only spheres A and B:
Charges will be induced on inner and outer surface of B to make the potential inside B equal at every point, thus making
Einside B=0. 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.

It's hard to find words that distinguish between the cavity inside the inner surface of a shell and the region between the surfaces of the shell. I understand that here you are referring to the latter, but it might not be clear. I use the term annulus to describe this.

You could further elaborate that the potential in the annulus due to induced charges on the inner surface exactly cancel the potential due to A, while those on the outer surface produce the constant potential kq/2a.

 

[SammyS]

How?I mean electric field lines of interior charge +q of A shell have to be terminated on inner surface of shell B as electric field can not exist in metal body hence electric field lines of charge +q of A shell would not able to go beyond inner surface of shell B.

I am not getting then how potential only depends on net charge present and not on induced charges .Isn't induced charge playing role equal to that of net charge present in determining potential?

The video addresses a situation very much like the situation you asked about in the following thread.

https://www.physicsforums.com/posts/5325447/
(A charge, q, external to a neutral conducting sphere. Asks for E-field and or potential due to induced charges.)​

The question asked in the current thread is not the same. The induced charges here have simple distributions. That's not the case for the video.

 

[gracy]

You mean there is potential due to induced charge but it's just that it does not exist in final formula(formula of net potential).When we have to calculate net potential we should ignore induced charge but when we are specifically asked potential due to induced charge we will have to consider induced charge then!
Here I am asked FINAL (net ) potential on shell B
Hence we will only consider charges which are actually present and will neglect induced charges
Suppose charge q' has been supplied by Earth as we know shell C has been grounded it's potential should be zero
##V_C##=##\frac{Kq}{3a}##+##\frac{Kq'}{3a}##=0

Therefore q'=-q

##V_B##=##\frac{Kq}{2a}## - ##\frac{Kq}{3a}##

=##\frac{Kq}{6a}##

Right?

 

[haruspex]

You mean there is potential due to induced charge but it's just that it does not exist in final formula(formula of net potential).When we have to calculate net potential we should ignore induced charge but when we are specifically asked potential due to induced charge we will have to consider induced charge then!
Here I am asked FINAL (net ) potential on shell B
Hence we will only consider charges which are actually present and will neglect induced charges

Not sure which post you are responding to, but nobody said that. All charges, induced or otherwise, result in potentials. Some induced charges may neutralise potentials due to other charges. In the case of the charges induced on the inner and outer surfaces of B, they neutralise each other (everywhere except in the annulus between them).

Suppose charge q' has been supplied by Earth as we know shell C has been grounded it's potential should be zero
VCV_C=Kq3a\frac{Kq}{3a}+Kq′3a\frac{Kq'}{3a}=0

Therefore q'=-q

VBV_B=Kq2a\frac{Kq}{2a} - Kq3a\frac{Kq}{3a}

=Kq6a\frac{Kq}{6a}

Right?

Yes.

 

[cnh1995]

You mean there is potential due to induced charge but it's just that it does not exist in final formula(formula of net potential).When we have to calculate net potential we should ignore induced charge but when we are specifically asked potential due to induced charge we will have to consider induced charge then!
Here I am asked FINAL (net ) potential on shell B
Hence we will only consider charges which are actually present and will neglect induced charges
Suppose charge q' has been supplied by Earth as we know shell C has been grounded it's potential should be zero
##V_C##=##\frac{Kq}{3a}##+##\frac{Kq'}{3a}##=0

Therefore q'=-q

##V_B##=##\frac{Kq}{2a}## - ##\frac{Kq}{3a}##

=##\frac{Kq}{6a}##

Right?

As haruspex said, we don't "neglect" the potential due to the induced charges. But since their contribution varies at each point, there isn't a general formula( like kq/r for net charge). Hence, only the net charge appears in the formula since it decides the net potential. In the net potential kq/r, potential due to induced charges is already included,hence, there's no need to calculate it separately if the "net potential" is asked. But to calculate the contribution of induced charges, as you said, you should consider the induced charges and the geometry of the system. Seeing the math you've done,
I believe you've understood the concept very well now!

 

[gracy]

You mean there is potential due to induced charge but it's just that it does not exist in final formula(formula of net potential).When we have to calculate net potential we should ignore induced charge but when we are specifically asked potential due to induced charge we will have to consider induced charge then!
Here I am asked FINAL (net ) potential on shell B
Hence we will only consider charges which are actually present and will neglect induced charges
Suppose charge q' has been supplied by Earth as we know shell C has been grounded it's potential should be zero

Which sentence is wrong here?

 

[SammyS]

Which sentence is wrong here?

Which post are you responding to ?

 

[gracy]

I have quoted .

 

[SammyS]

I have quoted .

You quoted yourself, but your question implies that you have been told that something in the quoted statement is wrong and you are responding the person who told you it's wrong.

So I say it in a slightly different way; Who are you addressing in the following?

Which sentence is wrong here?

 

[gracy]

Not sure which post you are responding to, but nobody said that

I thought this imply I was wrong somewhere.

 

[cnh1995]

Which sentence is wrong here?

I would say ,from mathematical point of view, nothing is wrong. In the final mathematical formula, only the net charge appears and induced charges don't, as you said. But physically, they are not "neglected". Their contribution is already "included" or "considered" in the net potential, which is varying from point to point,whose value is decided by the net charge. I think "neglect" and "ignore" are the words haruspex had problem with.

 

[gracy]

I think "neglect" and "ignore" are the words haruspex had problem with.

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

 

[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}##