# Charge inside

flyingpig

## Homework Statement

[PLAIN]http://img232.imageshack.us/img232/9330/11213091.jpg [Broken]

What is the E field in c < r < b?

Book say it is 0, but I don't agree

## The Attempt at a Solution

[PLAIN]http://img14.imageshack.us/img14/2926/42898284.jpg [Broken]

Let me explain, the blue circles are is the circle with radius r in c < r < b

Clearly, there is a E field from the solid sphere that's sending off its E-field upon entering the shell of the the bigger sphere and combing with the charges on the surface of the hollow sphere and exiting with a charge of 2Q

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Homework Helper
I understand what you did, but you are missing an important fact. The charges on the metal shell are free to move about. Electrons from the metal atoms will be pulled to the inner surface of the shell, terminating all the electric field lines coming from the positive central charge. A charge -Q will be on the inside surface of the shell. So the total charge inside your blue sphere will be zero.

flyingpig
I understand what you did, but you are missing an important fact. The charges on the metal shell are free to move about. Electrons from the metal atoms will be pulled to the inner surface of the shell, terminating all the electric field lines coming from the positive central charge. A charge -Q will be on the inside surface of the shell. So the total charge inside your blue sphere will be zero.

Can you draw a picture? I thought the conductor will force all the charges on the surface, I am not understanding you correctly.

flyingpig
[PLAIN]http://img29.imageshack.us/img29/2926/42898284.jpg [Broken]

This is what you mean right? If so, then how can the E field get out? Because at r = far away, the total Q will be 2Q

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Homework Helper
The point is the E field does not get out into the metal shell: E = 0 there as it must be inside any conductor. However, taking -Q out of the metal leaves +Q behind, and the positive charge pushes itself to the outer surface of the metal shell, adding to the +Q initially there. Electric field lines will go out from there. If you put an imaginary spherical surface outside of that, it will enclose a charge and give you an E field.

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Can you draw a picture? I thought the conductor will force all the charges on the surface, I am not understanding you correctly.
You have asked a question, only hinting at some of the IMPORTANT details.

What shapes are the objects (other than having circular cross-section)?

What materials are the object made of - insulators or conductors?

Any other details we should know?

flyingpig
What shapes are the objects (other than having circular cross-section)?

Spheres
What materials are the object made of - insulators or conductors?

Conductor, otherwise it wouldn't matter right?

flyingpig
Electric field lines will go out from there.

How can they go out if the lines keep canceling out each other?

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Spheres

Conductor, otherwise it wouldn't matter right?

Of course it matters.

Here's the deal with them being conductors.

The only way to maintain an electric field in the interior (a location surrounded by conducting material) of an electric conductor is to keep removing charges from one region an supplying them to another. In a static situation, this is NOT what is happening.

The electric field in the region: b < r < c, has and electric field of ZERO! If you consider a closed surface (such as a sphere with radius R : b < R < c), the electric flux through this surface is zero because the electric field is zero everywhere on the surface of this sphere. Therefore the net charge within the sphere of radius R is zero (Gauss's Law). Also, no net charge can reside (statically) at any location interior to a conductor. Therefore, there must be a net charge of ‒Q distributed on the inner surface of the conducting spherical shell with inner radius b. This is where your E field lines should terminate in your sketches. (If the spheres are concentric, this charge is distributed uniformly on the inner surface.)

If the net charge on the spherical shell is Q, and a charge of ‒Q is on the inner surface, what is the net charge on the exterior surface (radius = c) of the shell?

BTW: If there are no sources or sinks of electric field external to the spheres, then the charges on the exterior of the shell are distributed uniformly whether or not the solid sphere is at the center of the shell.

flyingpig
The electric field in the region: b < r < c, has and electric field of ZERO! If you consider a closed surface (such as a sphere with radius R : b < R < c), the electric flux through this surface is zero because the electric field is zero everywhere on the surface of this sphere.

Not according to my picture...

Therefore the net charge within the sphere of radius R is zero (Gauss's Law). Also, no net charge can reside (statically) at any location interior to a conductor. Therefore, there must be a net charge of ‒Q distributed on the inner surface of the conducting spherical shell with inner radius b. This is where your E field lines should terminate in your sketches. (If the spheres are concentric, this charge is distributed uniformly on the inner surface.)

If it terminates then how can it go out?

Homework Helper
E doesn't go out through the metal shell. It starts again on the outside of the shell. Note the increased charge there, allowing you to get the given answer for the field strength outside.

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Not according to my picture...
Then, your picture is wrong. Read my entire post with the idea of trying to understand what is being explained.

If it terminates then how can it go out?

The outer surface of the shell has a charge of +2Q. A whole different set of electric field lines originates on these charges.

flyingpig
The electric field in the region: b < r < c, has and electric field of ZERO!

If you consider a closed surface (such as a sphere with radius R : b < R < c), the electric flux through this surface is zero because the electric field is zero everywhere on the surface of this sphere.

But WHYY? My picture shows clearing the sphere inside is spreading its E-field

so, no net charge can reside (statically) at any location interior to a conductor

Why? other than telling me that the Efield inside is 0

Therefore, there must be a net charge of ‒Q distributed on the inner surface of the conducting spherical shell with inner radius b. This is where your E field lines should terminate in your sketches

I think you are leading me to say +Q + (-Q) = 0, but what about the inside? Because the net charge is +2Q outside of c

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The electric field in the region: b < r < c, has an electric field of ZERO!
(This should say: The electric field in the region: b < r < c, is ZERO!
A conductor is a material in which electric charge (usually negative charge) is relatively free to move. If there were to be a non-zero electric field within conducting material, electric charges in the material would be subject to electric force. Those charges free to move would move - and move in such a direction so that when they reach the boundary of the material they would then produce a layer of unbalanced charge which would produce an electric field opposed to the initial field. In the end, there cannot be a residual electric field within conducting material itself.

BTW: Early on in this thread, I asked whether the materials involved here were insulators or conductors. Your reply, "Conductor, otherwise it wouldn't matter right?" lead me to believe that you understood this very basic characteristic of a conductor.
If you consider a closed surface (such as a sphere with radius R : b < R < c), the electric flux through this surface is zero because the electric field is zero everywhere on the surface of this sphere.
But WHY? My picture shows clearing the sphere inside is spreading its E-field
That sphere is in a region with an electric field of zero.
Also, no net charge can reside (statically) at any location interior to a conductor.
Why? other than telling me that the E field inside is 0
I hope I've now answered this.

Therefore, there must be a net charge of ‒Q distributed on the inner surface of the conducting spherical shell with inner radius b.
I think you are leading me to say +Q + (-Q) = 0, but what about the inside? Because the net charge is +2Q outside of c
Qinner = ‒Q
Qshell = +Q
Also, Qshell = Qinner + Qouter
→   +Q = ‒Q + Qouter

You might rather think of it the situation in the shell in the following way:
Think of the inner sphere of as trying to produce an E field within the shell's material. Electrons in the shell will move toward the inner surface of the shell and away from the outer surface of the shell, until the excess negative charge on the inner surface and excess positive charge remaining on the outer surface produce an E field exactly opposite the field produced by inner sphere, leaving a net E field of zero.​

flyingpig
Here let's try it this way

http://www.electron.rmutphysics.com...rs-Serway-Beichne 6edr-4/24 - Gauss's Law.pdf

Go to page 13

There is no flux through the flat face of the cylinder inside the conductor because here E = 0

can you explain that?

I also don't understand where the Integral part, I thought they said E = 0 on the top and bottom (the quote above) and the other E is parallel.

In the example below (kinda the same one, but insulating)

Also I think I got it. The thing is there really is a E field in b < r < c, but it also cancels out with with the E field with the sphere.

[PLAIN]http://img209.imageshack.us/img209/9762/46808202.jpg [Broken]

This is as far as I have gotten because I don't know how that +Q is going to act...I don't how to make it go in the other direction

Sorry let me explain a bit again. They say the concentric larger circle (the one with negatives on one side and positive on the other) has a E field of 0 (E3 which I forgot to label), but there is a positive charge insulating sphere inside.

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flyingpig
E doesn't go out through the metal shell. It starts again on the outside of the shell. Note the increased charge there, allowing you to get the given answer for the field strength outside.

But how can it be 0 and suddenly add up outside.

Qinner = ‒Q
Qshell = +Q
Also, Qshell = Qinner + Qouter
→   +Q = ‒Q + Qouter

But why is Qinner = -Q? The sphere inside is +Q

You might rather think of it the situation in the shell in the following way:
Think of the inner sphere of as trying to produce an E field within the shell's material. Electrons in the shell will move toward the inner surface of the shell and away from the outer surface of the shell, until the excess negative charge on the inner surface and excess positive charge remaining on the outer surface produce an E field exactly opposite the field produced by inner sphere, leaving a net E field of zero.​

Yes, that hit my head now, but if we are far away and construct a Gaussian Surface over the whole thing, how will still that be +2Q? Because now that in some region it suddenly become 0, how can the outside suddenly be so big?

flyingpig

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Hello flyingpig:
(I'm having trouble getting this web site to get a quote from the right post.)

I said: "
Qinner = ‒Q
Qshell = +Q
Also, Qshell = Qinner + Qouter
→   +Q = ‒Q + Qouter "
You said: " But why is Qinner = -Q? The sphere inside is +Q "

You have given one part of the answer; that the charge of the inside sphere (radius a) is +Q. The other part is that the electric field in the (conducting) material of the spherical shell is zero. Apply Gauss's Law using any sphere whose surface is within the shell and has radius between b and c. This spere will have a NET charge of zero within its interior. The net charge is zero because the electric field is zero everywhere on this sphere's surface, so that there is no electric flux passing through its surface.

Gauss's Law says: $$\oint\vec{E}\cdot d\vec{A}=\frac{q_{in}}{\epsilon_0}$$

so that $$\oint\vec{E}\cdot d\vec{A}=0\quad\to\quad q_{in}=0$$

For this sphere, qin = Q plus the net charge on the inner sphere (Qinner).

qin = Q + Qinner → 0 = Q + Qinner → ‒Q = Qinner

Your last response & question seem to indicate that you are confusing electric charge and electric field.

" Yes, that hit my head now, but if we are far away and construct a Gaussian Surface over the whole thing, how will still that be +2Q? Because now that in some region it suddenly become 0, how can the outside suddenly be so big? "

What is it which "suddenly becomes so big" on the outside?

flyingpig
It's in the part where it talks about conductors and electrostatic equilibrium

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The first two properties of a conductor as listed by Serway, are important for this discussion.
1. The electric field is zero everywhere inside the conductor.
2. If an isolated conductor carries a charge, the charge resides on its surface.​

Serway does a good job of explaining these.

flyingpig
I am using Serway's book

flyingpig
For this sphere, qin = Q plus the net charge on the inner sphere (Qinner).

qin = Q + Qinner → 0 = Q + Qinner → ‒Q = Qinner

Because 0 = qin + qinside

0 = Q + (-Q)?

Your last response & question seem to indicate that you are confusing electric charge and electric field.

yes, I was, but I think I am over it now

" Yes, that hit my head now, but if we are far away and construct a Gaussian Surface over the whole thing, how will still that be +2Q? Because now that in some region it suddenly become 0, how can the outside suddenly be so big? "

What is it which "suddenly becomes so big" on the outside?

The E-field, it is 0 inside some inner region.

What is wrong with my picture?

[PLAIN]http://img209.imageshack.us/img209/9762/46808202.jpg [Broken]

The negative charges are on the surface and the positive charges on the inner surface. The sphere inside (insulating) is +Q, so how can the Field inside the conductor be 0? The two positive charges will have fields emerging from itself and touch the negative charge.

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flyingpig
I just realized a fundamental problem. The picture there cannot happen at anytime because the inner +Q will want to repel the sphere +Q and therefore the E-field is 0.

But what if I just have a conductor, if there is no external E-field, how can the inside of the conductor be 0?

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I just realized a fundamental problem. The picture there cannot happen at anytime because the inner +Q will want to repel the sphere +Q and therefore the E-field is 0.

But what if I just have a conductor, if there is no external E-field, how can the inside of the conductor be 0?
It's a bit hard to interpret your questions.

If you mean, what happens if we have a conducting spherical shell with charge +Q, and nothing in its interior (no sphere with charge +Q at the center), then the charges on the sphere will repel each other. The charge is free to move, so the charge will achieve equilibrium when it is distributed uniformly on the outer surface of the sphere, and only on the outer surface. That's also the only configuration for which the electric field in the conducting material of the shell is zero.

flyingpig
It's a bit hard to interpret your questions.

If you mean, what happens if we have a conducting spherical shell with charge +Q, and nothing in its interior (no sphere with charge +Q at the center), then the charges on the sphere will repel each other. The charge is free to move, so the charge will achieve equilibrium when it is distributed uniformly on the outer surface of the sphere, and only on the outer surface. That's also the only configuration for which the electric field in the conducting material of the shell is zero.

I mean by itself, let's say the positive charges are on the surface and the negatives are in the inner surface.

Also, could you show me that Qin = Qinner thing again? Can you use Qenclosed instead of Qinside?

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I mean by itself, let's say the positive charges are on the surface and the negatives are in the inner surface.

Also, could you show me that Qin = Qinner thing again? Can you use Qenclosed instead of Qinside?
What we call it is immaterial, as long as we're consistent.

So, which situation are you referring to?

I'll assume you mean the recent one with no sphere inside the shell.

If you initially place positive charge on the outer surface and negative on the inner, they will attract each other, and cancel each other out leaving the excess charge on the outer surface.

(Which post is "that Qin = Qinner thing" in? What is the number on that post - or is it in your other thread?)

flyingpig
If you initially place positive charge on the outer surface and negative on the inner, they will attract each other, and cancel each other out leaving the excess charge on the outer surface.
What do you cancel each other out?? There is no external E-field to cancel it out.

Let's just say I make a Gaussian Surface inside the conductor, I know the charge inside is 0 since the positive and negatives are distributed "up" and "below", but zero flux does not mean zero field, how do I determine?

flyingpig
Help?