Charge inside a spherical shell

[Tanya Sharma]

Homework Statement

A thin ,metallic spherical shell contains a charge Q on it. A point charge q is placed at the center of the shell and another charge q1 is placed outside it as shown in the figure .All the three charges are positive.

Q.1 The force on the charge at the center is

a) towards left
b) towards right
c) upward
d) zero

Ans : d) zero

Q.2 The force on the central charge due to the shell is

a) towards left
b) towards right
c) upward
d) zero

Ans : b) towards right

Homework Equations

The Attempt at a Solution

According to Gauss Law , charge q will induce charge –q on the inner surface of the shell ,so that charge on the outer surface of the shell is Q+q .

Central charge q will be under the influence of electric field due to the charge –q on the inner surface of the shell .By symmetry ,the force should be zero .

I feel the answer should be zero in both the cases . i.e option d) in both the cases.

How will the outer charge q1 affect the central charge inside the shell?

 

[haruspex]

According to Gauss Law , charge q will induce charge –q on the inner surface of the shell ,so that charge on the outer surface of the shell is Q+q .

Central charge q will be under the influence of electric field due to the charge –q on the inner surface of the shell .By symmetry ,the force should be zero .

The charge q1 will also induce charges on the shell. These will not be uniform, but will be so as to 'hide' q1 from being felt inside the shell. That is the reason there will be no force on q.
So, there's no force on q with both the shell and q1 present. What would the force have been on q if the shell were not there? What does that tell you about the force on q due to the charge on the shell?

 

[Tanya Sharma]

haruspex...thank you for the response

What would the force have been on q if the shell were not there?

The two charges will repel each other.q will feel a force towards left.

The charge q1 will also induce charges on the shell. These will not be uniform, but will be so as to 'hide' q1 from being felt inside the shell. That is the reason there will be no force on q.
So, there's no force on q with both the shell and q1 present.

Were the outside charge q1 not present ,the central charge would not have felt any force due to spherical shell.

Now when outside charge q1 is present ,the net force on the central charge is still zero .

Is there any principal that a force on a charge inside a cavity has to be zero ??

 

[Tanya Sharma]

Why does the force on a charge inside a cavity has to be zero ??

 

[haruspex]

Why does the force on a charge inside a cavity has to be zero ??

I'm not saying it would be in general, but it is in this case. Since the shell is a conductor, it will all be at one potential. The field inside the shell depends only on that and any charges inside the shell. Since the only charge inside is central, the field is symmetric about the centre. Hence no force on the charge.

 

[Tanya Sharma]

Does the force on the central charge q due to outer charge q1 remains the same irrespective of whether the spherical shell is present or not ?

 

[haruspex]

No. As I explained above, it shields q from any awareness of q1. Its charge distribution therefore provides a force exactly cancelling that from q1.

 

[Tanya Sharma]

No. As I explained above, it shields q from any awareness of q1. Its charge distribution therefore provides a force exactly cancelling that from q1.

This is the point I am unable to comprehend...I understand how the shell is shielding but why does the shell shields the central charge ?

 

[aralbrec]

This is the point I am unable to comprehend...I understand how the shell is shielding but why does the shell shields the central charge ?

According to Gauss Law , charge q will induce charge –q on the inner surface of the shell ,so that charge on the outer surface of the shell is Q+q .

This is correct and is found with Gaussian surface in the interior of the conducting shell.

Central charge q will be under the influence of electric field due to the charge –q on the inner surface of the shell .By symmetry ,the force should be zero .

However you cannot draw this conclusion by argument of symmetry. The central charge is being acted on by all charges in space, including the charge induced on the inner and outer part of the shell and the charge outside the shell.

When the outer charge is not present, you *can* draw the symmetry conclusion. This is because the charge on the outer and inner surfaces of the shell will distribute evenly. This must happen, otherwise there is a tangential field on the surfaces of the shell that will cause surface charge to continue to move. So in the static case, the surface charges must evenly distribute. Then you can look at the interior of the shell and state the field must be directed toward the center of the shell by symmetry because of the even distribution of charge on the shell surfaces (there is no preferential location inside the shell as a function of theta). Only then can you use Gauss's Law to find the interior field strength. Gauss's Law cannot be fruitfully applied until you can make this symmetry argument. It is important to note this: all charges on the outer surface of the shell and the inner surface of the shell are acting on the point charge in the interior, however their distribution causes the field in the interior to be directed radially toward the shell center. This allows Gauss's Law to be applied fruitfully to find the field strength. I think I am rambling here, but it is to get to this point that many students misunderstand: Gauss's Law does not state that the field on its surface is only generated by charge interior to its surface -- this is absolutely incorrect. The field at any point in space is determined by the distribution of charges in all space. In this example, those charges outside the Gaussian surface affect the field inside the shell by controlling the field direction, making it radial.

In the current example with a charge also placed outside the shell, charge on the outer shell will not be distributed evenly -- a higher concentration of charge on the outer shell will appear nearest the outside charge and, due to conservation of charge, a lower concentration of charge will appear on the side of the outer shell diametrically opposite. The symmetry argument is gone.

Instead, as haruspex points out, you can take advantage of the fact the shell is a perfect conductor and therefore sits at the same potential throughout. So in the interior of the shell we can rely on Laplace's Equation ∇²V = p_v / ε. This equation says the field on the inside of the shell depends only on the charge distribution inside the shell and the voltage on the boundary of the inner shell. Since the voltage is everywhere the same on the inner shell, the field must be radially directed to the center and again Gauss's Law can be used to find the field inside the shell. And you find the field is independent of any outside charge.

Since you know the field inside the shell is the same whether or not the outside charge exists, you know that every charge in space exerts a force on the charge inside the shell, you know exactly the force the outside point charge is exerting on the inside point charge (coulomb's law), what kind of force must the charge distribution on the outer shell be exerting on the inner point charge in order for it to experience a net force of zero?

 

[ehild]

I would not ask such questions, as the force the shell exerts on the charge. That force cannot be measured by any way.

The charge inside is screened from the effect of any outer charge by the closed metal shell. You can not calculate with Coulomb force between q and q1. q and q1 do not interact as they are screened from each other.

Coulomb equation F=kQ₁Q₂/r² with k=1/(4πε₀) is valid if the point charges are in vacuum. In an other material, the force is smaller, as ε₀ is replaced by ε, the dielectric constant of the material. If you put any material between two point charges, you can not calculate the force between them with the Coulomb equation.

The central charge inside interacts with the charges on the inner surface. The surface charge is bounded to the metal shell, it can not move away from it, so we can imagine force of interaction between the surface charges and shell. This way a tiny piece of the surface experiences some force from the central charge, but the net force is zero because of symmetry. The shell experiences zero net force from the central charge, so the central charge experiences zero force from the shell.

It is an other thing that you can replace the metal shell by the surface charge density on both surfaces, as if the metal was not present, when calculating electric field. (For that you have to know the charge distribution.) In this way, you can apply Coulomb force as if all the charges were in vacuum and assign a force between the inner central charge and the outer surface charge distribution. But that is not a real, measurable force.

ehild

 

[Tanya Sharma]

The central charge is being acted on by all charges in space, including the charge induced on the inner and outer part of the shell and the charge outside the shell.

It is important to note this: all charges on the outer surface of the shell and the inner surface of the shell are acting on the point charge in the interior, however their distribution causes the field in the interior to be directed radially toward the shell center

Do the charge on outer surface and the charge q1 affect the electric field inside the shell ? If yes , then how is the shell shielding the inner charge q .

In the current example with a charge also placed outside the shell, charge on the outer shell will not be distributed evenly -- a higher concentration of charge on the outer shell will appear nearest the outside charge and, due to conservation of charge, a lower concentration of charge will appear on the side of the outer shell diametrically opposite.

I think it should be the other way ,lower concentration on the surface near the outside charge and vica versa .

This equation says the field on the inside of the shell depends only on the charge distribution inside the shell and the voltage on the boundary of the inner shell. Since the voltage is everywhere the same on the inner shell, the field must be radially directed to the center and again Gauss's Law can be used to find the field inside the shell. And you find the field is independent of any outside charge.

Since you know the field inside the shell is the same whether or not the outside charge exists, you know that every charge in space exerts a force on the charge inside the shell, you know exactly the force the outside point charge is exerting on the inside point charge (coulomb's law)

Isnt it a contradiction that the central charge q is under the influence of electric field due to induced charge on the inner surface which remains the same irrespective of presence outside charge q1 and at the same time the central charge is being acted on by force from charge on outer shell and the external outside charge q1 ?

It is an other thing that you can replace the metal shell by the surface charge density on both surfaces, as if the metal was not present, when calculating electric field. (For that you have to know the charge distribution.)

Is the metal or the charge layer resposible for shielding the inner charge from outside charge ? Forgive me if this question sounds stupid .

The inner central charge is under the influence of electric field form the induced charges on the inner surface of the shell .Does the electric field inside the shell change in the presence of the outside charge q1 ?

 

[ehild]

Is the metal or the charge layer responsible for shielding the inner charge from outside charge ? Forgive me if this question sounds stupid .

The metal shell shields the inner charge from any fields of the charges present outside it.

The inner central charge is under the influence of electric field form the induced charges on the inner surface of the shell .Does the electric field inside the shell change in the presence of the outside charge q1 ?

No.

 

[Tanya Sharma]

ehild...Thanks!

So ,the central charge is under the influence of electric field from the induced charges on the inner surface of the shell , and this electric field doesn't change in this question .The central charge experiences any force due to this field , which remains unchanging in our case .

So,the force from the shell on the central charge should be zero and the force of the outside charge q1 on q should also be zero .Isnt it ??

 

[ehild]

I do not think "what is the force from the shell" a good question. The central charge interacts with the inner surface charges on the metal shell, not with the shell itself. That force is zero. The outer surface charge does not act to the central charge.

ehild

 

[Tanya Sharma]

I do not think "what is the force from the shell" a good question. The central charge interacts with the inner surface charges on the metal shell, not with the shell itself. That force is zero. The outer surface charge does not act to the central charge.

ehild

Yes...your objection to the question sounds reasonable...

The presence of outer charge merely affects the charge distribution on the outer surface of the shell ...fewer positive charges towards the outer charge q1 and more positive charge towards right.ie away from q1 .Isnt it ??

 

[ehild]

Yes...your objection to the question sounds reasonable...

The presence of outer charge merely affects the charge distribution on the outer surface of the shell ...fewer positive charges towards the outer charge q1 and more positive charge towards right.ie away from q1 .Isnt it ??

Yes, it is :) (but is not it to the left away from q1?)
ehild

 

[Tanya Sharma]

Yes, it is :) (but is not it to the left away from q1?)



ehild

Sorry..I replied hastily

Thank you very much...You are fabulous...

 

[Tanya Sharma]

Hello ehild !

It is an other thing that you can replace the metal shell by the surface charge density on both surfaces, as if the metal was not present, when calculating electric field. (For that you have to know the charge distribution.) In this way, you can apply Coulomb force as if all the charges were in vacuum and assign a force between the inner central charge and the outer surface charge distribution. But that is not a real, measurable force.

ehild

Isn't this a contradiction to the fact that the central charge is under the influence of the electric field due to only the induced charges on the inner surface of the shell and not from the outer charge q1 as well the charges on the outer surface ? Since the central charge is not interacting with the outer charge q1 and also not with the charges on the outer surface of the shell ,so while calculating the electric field at the centre of the shell , shouldn't we consider only the induced charges on the inner surface of the shell ?

 

[haruspex]

the central charge is under the influence of the electric field due to only the induced charges on the inner surface of the shell and not from the outer charge q1 as well the charges on the outer surface.

No, that's not correct. You can look at it in two different ways
1. The conducting shell shields the charges inside it from those outside in the sense that the field inside the shell is the same as if the charges outside were not there. (But note that if the charges outside were not there then the charge on the shell would be distributed differently.)
2. The field from charges outside the shell permeates the shell, but the induced change in distribution on the shell creates a field that exactly cancels it.
So you can include both the external charges and the induced charge, or exclude both.

 

[Tanya Sharma]

Hello haruspex

The field from charges outside the shell permeates the shell, but the induced change in distribution on the shell creates a field that exactly cancels it.
So you can include both the external charges and the induced charge, or exclude both.

By induced charges you mean charges on the outer surface of the shell or inner surface or both?

 

[haruspex]

Hello haruspex



By induced charges you mean charges on the outer surface of the shell or inner surface or both?

Wherever

 

[Tanya Sharma]

So we are applying the two concepts simultaneously.

1. Every charge in space exerts a force on the charge inside the shell.
The central charge is being acted on by all charges in space, including the charge induced on the inner and outer part of the shell and the charge outside the shell.

2. But at the same time field inside the shell is the same whether or not the outside charge exists and is determined solely by the charges induced on the inner surface of the shell .Now due to spherical symmetry of the inner surface of the shell the net electric field is zero at the center of the shell .So ,with all three charge distribution ,inner surface charge, outer surface charge, charge q1 acting on the central charge ,the central charge experiences zero force .

Is my thinking correct?

 

[ehild]

No, you can not apply the two concepts simultaneously.

The method which includes all charges and induced charges, replaces the metal shell with two spherical layers of charges. The electric field at a point is the sum of the contributions of all charges. There are dq/ε₀ field lines emerging or ending from a surface element with charge dq. You know that the field is zero in the space between the spherical surfaces. It is not easy to calculate the contributions as you do not know the induced charge distribution on the outer surface.

ehild

 

[Tanya Sharma]

Hello ehild

"Every charge in space exerts a force on the charge inside the shell.
The central charge is being acted on by all charges in space, including the charge induced on the inner and outer part of the shell and the charge outside the shell."

Is this correct ?

 

[haruspex]

So we are applying the two concepts simultaneously.

1. Every charge in space exerts a force on the charge inside the shell.
The central charge is being acted on by all charges in space, including the charge induced on the inner and outer part of the shell and the charge outside the shell.

2. But at the same time field inside the shell is the same whether or not the outside charge exists and is determined solely by the charges induced on the inner surface of the shell

I'd word the second differently:
As a result the field inside the shell is the same whether or not the outside charge exists; the induced charge exactly cancels them. The field inside is determined solely by the charges inside the shell.

Now due to spherical symmetry of the inner surface of the shell the net electric field is zero at the center of the shell .

It's not to do with symmetry. The charges in a conductor necessarily move so as to arrange that the entire conductor is at one potential. If there are no other charges inside the shell then that potential will apply throughout the cavity, so no internal field.

 

[SammyS]

Hello ehild

"Every charge in space exerts a force on the charge inside the shell.
The central charge is being acted on by all charges in space, including the charge induced on the inner and outer part of the shell and the charge outside the shell."

Is this correct ?

Yes, this is true , but --- (That's the dreaded "Oh! Oh! here it comes now", but) ---

Consider the electric field due only to the external charge and the charge on the outer surface of the sphere:

The external charge induces the charges on the exterior of the conductor (in this case, the sphere) to distribute themselves in such a way that the electric field inside the surface of the conductor is zero.

The net effect of this is that no matter where you place charges external to the conductor the field produced (by these charges plus the surface charge) at any location within the outer surface of the conductor can be ignored, because that field is zero. Of course, this assumes the external charges are stationary.​

Any field inside the inner surface of the conductor is due to the charges on the inner surface and any charges within the conductor.

 

[Tanya Sharma]

haruspex...SammyS...ehild...thank you very much