Variation of Electric Field at the centre of Spherical Shell

[Shreya]

Homework Statement

A charge Q is uniformly distributed on a thin spherical shell. If a point charge is brought close to the shell, will the field at the centre change? Does your answer depend on whether the shell is conducting or not?

Relevant Equations

Gauss Law: Surface Integral of E*dA = q(enclosed)/Eplison
* - dot product

My approach is thus: the shell will have induced charges if it's conducting resulting in E at the centre of shell(though flux at centre will be 0). For non conducting spheres there can be no induction only polarization of dipoles, therefore the E field at centre will remain 0. Is my approach correct?
Please be kind to help

 

[ergospherical]

For non conducting spheres there can be no induction... therefore the E field at centre will remain 0.

That's true for the electric field of the shell, but what about the electric field from the external charge? Remember the principle of superposition!

My approach is thus: the shell will have induced charges if it's conducting resulting in E at the centre of shell

Again, you're neglecting the field of the external charge. But this conducting case is somewhat more subtle - have you ever come across the idea of a Faraday cage?

It's a result of electrostatics that the electric field inside a hollow conductor is zero, which follows from the uniqueness theorem (i.e. you're solving $\nabla^2 \phi = 0$ with $\phi = \phi_0$ on the boundary, which has a unique solution of $\phi = \phi_0$ throughout).

 

[Shreya]

have you ever come across the idea of a Faraday cage?

Yes, I have! So the E inside the shell will always be zero, right? (There will be no induced charges inside)

About the nonconducting spheres, do you mean that the E is present due to the polarization of dipoles in the non conducting sphere?

 

[ergospherical]

Yes, I have! So the E inside the shell will always be zero, right?

The electric field inside the hollow conducting shell will be zero, yeah!

If you want to look at it in terms of charge distributions (instead of via Laplace's equation): anywhere inside the shell, the electric field of the external charge is exactly balanced by the electric field of the charge distribution induced in the conducting shell.

About the nonconducting spheres, do you mean that the E is present due to the polarization of dipoles in the non conducting sphere?

In the non-conducting shell, the charges on the sphere are fixed (and given to be distributed uniformly). By the shell theorem (or equivalently, by Gauss' law), the field inside due to the shell is zero.

But in accordance with the principle of superposition, you still need to take into account the field of the external charge. So the field will be non-zero inside the shell!

 

[Shreya]

The electric field inside the hollow conducting shell will be zero, yeah!

If you want to look at it in terms of charge distributions (instead of via Laplace's equation): anywhere inside the shell, the electric field of the external charge is exactly balanced by the electric field of the charge distribution induced in the conducting shell.In the non-conducting shell, the charges on the sphere are fixed (and given to be distributed uniformly). By the shell theorem (or equivalently, by Gauss' law), the field inside due to the shell is zero.

But in accordance with the principle of superposition, you still need to take into account the field of the external charge. So the field will be non-zero inside the shell!

Ah-ha! I got it! Thanks a lot ergospherical!

 

[jbriggs444]

polarization of dipoles in the non conducting sphere?

This strikes me as the next best thing to an oxymoron.

For any finite charge, and any finite potential difference, if you put enough dipoles into a "non-conducting" substance, it can conduct that much charge when that much potential difference is applied. [It's called capacitance].

 

[Steve4Physics]

For any finite charge, and any finite potential difference, if you put enough dipoles into a "non-conducting" substance, it can conduct that much charge when that much potential difference is applied. [It's called capacitance].

I suspect that @Shreya was considering the effects arising from induction in the insulator.

When the external point charge (say q) is brought close to the shell, induction occurs because of the production of dipoles (polarisation of the insulator’s atoms/molecules).
https://www.ux1.eiu.edu/~cfadd/1360/23EFields/23Images/Fig23.04a.jpg
This creates additional surface charge distributions (both inner and outer surfaces of the shell).

So we now have to add three fields to find the total at the centre:
a) the field from Q (which will be zero due to synmetry);
b) the field from q;
c) the field from the induced charges.

For a thick shell, the field from the induced charges is pretty tricky to determine. However the question specifically says a thin spherical shell. Providing the shell is thin enough, the field due to the induced charges will be negligible – I think that's what the question intends.

 

[Shreya]

However the question specifically says a thin spherical shell. Providing the shell is thin enough, the field due to the induced charges will be negligible – I think that's what the question intends.

Yeah, I never thought about that point. But even if the shell isn't thin enough the E inside is non zero, right?

Sorry for the awful diagram.

if you put enough dipoles into a "non-conducting

I know I wasn't clear enough.As @Steve4Physics said, I meant polarization of atoms resulting in dipoles

 

[Steve4Physics]

But even if the shell isn't thin enough the E inside is non zero, right?

Right. The induced field (at the centre of the non-onducting shell) is always non-zero. The shell thickness only affects the magnitude.

Note, your diagram shows the induced field to be uniform and to the left. But at each point in the hollow region, the induced field’s magnitude and direction depends on a number variables. The field has a more complex shape than a uniform field.

At the centre of the insulating shell, the induced field points in the same direction as the field from q; the total field at the centre (and probably elsewhere) is therefore larger than the external field from q alone.

 

[Shreya]

Note, your diagram shows the induced field to be uniform and to the left. But at each point in the hollow region, the induced field’s magnitude and direction depends on a number variables. The field has a more complex shape than a uniform field.

Yes, that's true
Thanks a lot @Steve4Physics! I got a lot of new insights!

 

[rudransh verma]

Homework Statement:: A charge Q is uniformly distributed on a thin spherical shell. If a point charge is brought close to the shell, will the field at the centre change? Does your answer depend on whether the shell is conducting or not?
Relevant Equations:: Gauss Law: Surface Integral of E*dA = q(enclosed)/Eplison
* - dot product

My approach is thus: the shell will have induced charges if it's conducting resulting in E at the centre of shell(though flux at centre will be 0). For non conducting spheres there can be no induction only polarization of dipoles, therefore the E field at centre will remain 0. Is my approach correct?
Please be kind to help

The charges in the conducting shell should be on the surface unlike in a non conducting shell where the charges are fixed all inside the shell. If it’s a conducting shell then due to presence of an external charge there should be induction and the charges should spread out on the outer and inner surface of the shell.
I am not sure what will be the pattern of field will be?

 

[Steve4Physics]

The charges in the conducting shell should be on the surface unlike in a non conducting shell where the charges are fixed all inside the shell.

That's not strictly true. You are ignoring the effects of induction by polarisation in the non-conducting shell (see Posts #7 - #10).

I am not sure what will be the pattern of field will be?

You can't work out the pattern unless you know the signs of Q and the external point charge. But, in fact, you do not need to know this to answer the original question (in Post #1)

 

[Shreya]

If it’s a conducting shell then due to presence of an external charge there should be induction and the charges should spread out on the outer and inner surface of the shell.

I think the charges will only be present on the outer surface (electrostatic shielding)

 

[rudransh verma]

“A charge Q is uniformly distributed on a thin spherical shell. If a point charge is brought close to the shell, will the field at the centre change?”

@Steve4Physics without the presence of an external charge the field inside the conducting/non conducting shell should be zero and outside perpendicular to the surface.
Now if it’s conducting then the charges should induce on inner surface but still the field inside should be zero. I don’t exactly know how will it be between q(external charge) and charges on outer surface.

The induced field (at the centre of the non-onducting shell) is always non-zero.

Shouldn’t it be zero by gauss law ? I am not familiar with induction by polarisation of dipoles in non conductor shell.

 

[Steve4Physics]

@Steve4Physics without the presence of an external charge the field inside the conducting/non conducting shell should be zero and outside perpendicular to the surface.

Agreed for the conducting shell. For a non-conducintg shell, only the field at the centre is zero (by symmetry); but the field increases as you move from the centre to the inner surface.
Edit: Whoa! Sorry, that’s wrong. In this case, the field is zero everywhere inside the shell.

Now if it’s conducting then the charges should induce on inner surface but still the field inside should be zero.

Agreed.

Shouldn’t it be zero by gauss law ? I am not familiar with induction by polarisation of dipoles in non conductor shell.

Gauss's law tells us that if a region contain zero charge, then there is net zero electric flux through the surface enclosing the region.

This does not mean a charge-free region has zero electric field. If there is (say) a point charge outside the region, then there is a field inside the region even though the region contains no charge.

Polarisation of the non-conducting shell will produce an additional internal electric field. Though understanding this is not required in order to answer the original question.

 

[rudransh verma]

For a non-conducintg shell, only the field at the centre is zero (by symmetry); but the field increases as you move from the centre to the inner surface.

For thin/thick conductor shell the charges will be at the outer surface. Field inside will be zero.
For thin non conducting shell field is again zero inside.
For thick non conducting shell idk ?

 

[Steve4Physics]

1. It is not clear if you mean with/without the external charge.

For thin/thick conductor shell the charges will be at the outer surface. Field inside will be zero.

2. Agree - with or witout external charge.

For thin non conducting shell field is again zero inside.

3. Disagree.

With no external charge present, the field is zero only at the centre. The field's magnitude inceases as you move radially outwards from the centre to the shell's inner surface (because of embedded charge Q).
Edit: Whoa! Sorry, that’s wrong. In this case, the field is zero everywhere inside the shell. So I agree.With the external charge present, the field inside is non-zero everywhere due to the induced field from the (assymetrical) polarisation.

For thick non conducting shell idk ?

4. Same as point 3, above. But when the external charge is present, the induced field will be larger than that of a thin shell. (Minor edit made.)

 

[rudransh verma]

With no external charge present, the field is zero only at the centre. The field's magnitude inceases as you move radially outwards from the centre to the shell's inner surface (because of embedded charge Q).

You mean like this. At center it’s symmetrical and off center it’s not?

 

[Steve4Physics]

@Shreya and @rudransh verma. My apologies. I got something wrong.

For an isolated non-conducting spherical shell, with charge Q uniformly distributed, the field everywhere inside the shell (not just at the centre) is zero. This follows from Gauss’s Law and symmetry.

Unfortunately, part of my brain had auto-tuned into thinking about a solid sphere of uniform charge density.

However, it is still correct to say that the external point charge makes the internal field (inside the shell's hollow) non-zero. This internal field is the sum of the field from the point charge and the field from the induced dipoles.

Apologies again for the confusion.

 

[rudransh verma]

My apologies. I got something wrong.

For an isolated non-conducting spherical shell, with charge Q uniformly distributed, the field everywhere inside the shell (not just at the centre) is zero. This follows from Gauss’s Law and symmetry.

Yeah! You confused me. I thought maybe I am missing something.
To summer rise :
In the absence of external charge field inside a thin/thick conducting/ nonconducting shell will be zero everywhere. This is true from Guass’s law.

But based on my drawing don’t you think you are right? That field only at the center should be zero not everywhere whether it’s a conductor or nonconductor shell, thick or thin?

 

[Steve4Physics]

Yeah! You confused me. I thought maybe I am missing something.
To summer rise:

Sorry! (And you mean 'summarise'!)

In the absence of external charge field inside a thin/thick conducting/ nonconducting shell will be zero everywhere. This is true from Guass’s law.

Agreed. But it's worth being aware of some implicit assumptions. The field is zero providing:

- there is electrostatic equilibrium (so no current is flowing through the conductor and no changing polarisation in the non-conductor);

- for the non-conductor, any existing charge is uniformly distributed (spherically symmetric).

If these conditions are not met, the field inside could be non-zero.

But based on my drawing don’t you think you are right? That field only at the center should be zero not everywhere whether it’s a conductor or nonconductor shell, thick or thin?

Your drawing refers to a completely different situation. You have put the point charge is inside the shell.

Also your field lines have different lengths. I don't know if that's intentional. Field lines are not vectors so different lengths have no meaning. (At any point on a field the electric field vector is a tangent to the field line.)

For information, here's a diagram for the field for a point charge inside a conducting shell: https://i.stack.imgur.com/CJgG9.png
I can't find a diagram for a non-conducting shell (edit: uncharged shell)

 

[Shreya]

For an isolated non-conducting spherical shell, with charge Q uniformly distributed, the field everywhere inside the shell (not just at the centre) is zero. This follows from Gauss’s Law and symmetry

Its alright! It happened to me too :)There's a subtle argument regarding this:
Please refer to the image.

In a generic case, there would be a an angle between Area and r vector, but the cos theta would cancel from both sides.
For a thick shell, you could assemble many concentric thin shells within.

 

[rudransh verma]

Your drawing refers to a completely different situation. You have put the point charge is inside the shell.

Also your field lines have different lengths. I don't know if that's intentional. Field lines are not vectors so different lengths have no meaning. (At any point on a field the electric field vector is a tangent to the field line.)

I didn’t put any charge inside. Just a point where I have shown the vectors. At center the field vectors will be equal . So field there should be zero. Off center vectors due to charge elements should be of unequal lengths and so a net field everywhere off center. But again that’s different from what gauss law tells(field is zero because flux is zero because no charge) .
Can you clarify?

 

[Shreya]

But again that’s different from what gauss law tells(field is zero because flux is zero because no charge) .
Can you clarify?

I think if my previous post has a clue for you. The E caused by Area1 is Cancelled out by E of A2. For every area element, you can find another one. Therefore, E will be 0Edit: the diagram is for a conducting shell.
For a non conducting shell: Charge will also reside in the inner surface. That would result in flux lines inside, yet the E is zero. Please correct me if I am wrong

 

[Steve4Physics]

I didn’t put any charge inside. Just a point where I have shown the vectors. At center the field vectors will be equal . So field there should be zero. Off center vectors due to charge elements should be of unequal lengths and so a net field everywhere off center.

Oh I see! What you have drawn looks the same as field lines emanating from a point charge. And there was no accompanying explanation. Hence my misunderstanding.

Doing the maths, it turns out that the vector-sum of all field contributions at any internal point is zero. E.g. see @Shreya's Post #23. Interestingly, the 'zero field' is true for any shape hollow conductor (not just spherical) and is a consequence of the inverse square law.

But again that’s different from what gauss law tells(field is zero because flux is zero because no charge) .
Can you clarify?

If you apply Gauss's law, you have to specify what (closed) surface you are using. I can't tell what closed surface you are considering.

 

[rudransh verma]

Doing the maths, it turns out that the vector-sum of all field contributions at any internal point is zero. E.g. see @Shreya's Post #23.

I don’t get the math. Can you explain?

 

[Steve4Physics]

I don’t get the math. Can you explain?

It is the application of Newton's shell theorem to charge and electric fields. (The original shell theorem applies to mass and gravitational fields but works just the same for charge and electric fields.) There is plenty available if you do a search, e.g. https://en.wikipedia.org/wiki/Shell_theorem#Force_on_a_point_inside_a_hollow_sphere

 

[Shreya]

I don’t get the math. Can you explain?

If you want to: you can watch Lecture 4 from yale university Fundamentals of Physics II lectures (21:03 to 45:10).