Charge inside and outside conducting shell

[Utsavdutta98]

Homework Statement

A thin metallic spherical shell contains a charge Q over it. point charge +q is placed in side the shell at point T separated from the centre by a distance a. Another point charge q1 is placed outside the shell at a distance b from the centre find the electric field at the centre due to the charge over outer surface of the shell and the net field at the centre. (See figure)

Homework Equations

Gauss law. Coulombs law. We know that the field inside a conductor due to the superposition of fields due to external charges is equal to zero.

The Attempt at a Solution

There will be a field due to the charge inside the shell i.e. q. This charge will induce a non-uniform charge -q on inner surface with q+Q on outer surface. This charge q+Q will be spread non uniformly due to the presence of q1. So shouldn't the field caused by charges on outer surface be non-zero due to lack of any symmetry. And will the field of q1 exactly cancel out the field of q+Q. And if so, why?

 

[Simon Bridge]

Welcome to PF;
These are good questions - the charge over the shell will, indeed, be displaced by the presence of the other charges.
The trick is to work out how... what is the rule for how the charges in a conductor will distribute themselves?
Then use that rule in the case there is one charge outside the shell (what is the field strength at the center?) and repeat for a single charge inside the shell.
You may also want to read about Faraday cages.

 

[Utsavdutta98]

Thanks for the reply Simon! From what I know charges outside the conductor and on the surface arrange themselves in such a way that the field inside is 0. So i would think that the superposition of the fields due to Q+q (surface charge) and q1 should be zero. But that does not necessarily mean that the field due to Q+q will alone be zero.

 

[Simon Bridge]

The charges in a conductor will always rearrange so that the net force on each one is zero. This happens when there is a zero electric field inside the conductor - because that's what "zero electric field" means. Notice that's all the charges in a conductor. A neutral conductor has equal positive and negative charges - a charged conductor has a surplus.

What happens in the case that you have a single charge (not in the middle) inside a neutral conducting shell - what does the field inside and outside look like (can you sketch the lines of force?)

 

[Utsavdutta98]

From what I understand there will definitely be a non zero field at the centre due to the presence of an internal charge. What i can't understand is that will E(surface) + E(q1) = 0? The field lines are attached.

 

[Simon Bridge]

What i can't understand is that will E(surface) + E(q1) = 0? The field lines are attached.

If they are not - then the charges on the conductor will experience a non-zero force along the conductor and the situation is not longer static which contradicts the problem statement (context). Therefore - the electric fields in the conductor have to be zero. I do not understand why it's such a problem for you.

The charges in a conductor will always move to cancel out any electric field that would otherwise be there.

 

[Utsavdutta98]

So will the superposition of all three charges systems be 0 or will the superposition of only the external 2 charge systems be zero? And even if their sum is zero, which is true, then what can we comment about their INDIVIDUAL fields?

 

[vela]

Thanks for the reply Simon! From what I know charges outside the conductor and on the surface arrange themselves in such a way that the field inside is 0. So i would think that the superposition of the fields due to Q+q (surface charge) and q1 should be zero. But that does not necessarily mean that the field due to Q+q will alone be zero.

I'm kind of guessing what you're asking here. This is what I'm getting. When you have the fields due to the surface charge Q+q and the point charge q1, you can superpose the two fields so that they will cancel inside the conductor, but now if you remove q1 from the situation, then how can the surface charge Q+q result in a zero field inside the conductor because there's no longer a second field to cancel it? The answer is that the field due to every little piece of Q+q add up to cancel inside the conductor.

When q1 isn't present, the surface charge q+Q will distribute itself uniformly on the surface of the conductor. If you now calculate the total electric field anywhere inside the sphere of charge, you'll find it's equal to 0. This is completely analogous to the gravitational case where there's no gravitational field inside a thin spherical shell of mass. Imagine a test charge at some point in the interior. Consider an infinitesimal section of the sphere and the corresponding section on the opposite side of the sphere where both sections subtend the same solid angle. Each section has a charge that's proportional to the square of its distance from the point, so it turns out the force exerted by each section is independent of its distance from the point. Since the force point in opposite directions, they cancel. Integrating over the entire sphere, you can show that the force (and hence the field) is 0 everywhere inside the sphere.

 

[andrea96]

Homework Statement

A thin metallic spherical shell contains a charge Q over it. point charge +q is placed in side the shell at point T separated from the centre by a distance a. Another point charge q1 is placed outside the shell at a distance b from the centre find the electric field at the centre due to the charge over outer surface of the shell and the net field at the centre. (See figure)

Homework Equations

Gauss law. Coulombs law. We know that the field inside a conductor due to the superposition of fields due to external charges is equal to zero.

The Attempt at a Solution

There will be a field due to the charge inside the shell i.e. q. This charge will induce a non-uniform charge -q on inner surface with q+Q on outer surface. This charge q+Q will be spread non uniformly due to the presence of q1. So shouldn't the field caused by charges on outer surface be non-zero due to lack of any symmetry. And will the field of q1 exactly cancel out the field of q+Q. And if so, why?

For the net field is it possible that the answer is $$E=\frac {kq (R^3-a^3)}{a^2R^3}$$?

 

[Simon Bridge]

@andrea96: How did you get that result for the net field at the center of the spherical shell?

Notice that the center of the shell and the locations of the two charges do not have to be on the same line?
i.e. The exact positions are not stated. This is a clue.

Lynchpin: The net field, inside the sphere, due to all charges on and outside the sphere, is zero.

 

[andrea96]

@andrea96: How did you get that result for the net field at the center of the spherical shell?

Notice that the center of the shell and the locations of the two charges do not have to be on the same line?
i.e. The exact positions are not stated. This is a clue.

Lynchpin: The net field, inside the sphere, due to all charges on and outside the sphere, is zero.

Really I think the distribution on the inner surface doesn't depend on what happen on the outer surface and out of the spher: in fact I could divided the sphere with a circle line in the middle of the outer and inner surface without changing somethig and so I' d create two cincentric conductors; so the external conductor shield whatever is inside! I can treat the inside system indipentently by the external system and it is constitutes of a point charge q inside a spherical cavity of a spherical conductor with charge -q, and I have studied it using image method.
P. S. why you think the total field have to be zero? the field in a empty cavity of a conductor is always zero, but if in the cavity there is a free, charge the field isn't zero... think about the field inside a spherical concentric cavity of a sphere with a point charrge in the centre of the cavity...

 

[Simon Bridge]

P. S. why you think the total field have to be zero?

I don't. Please read more carefully:
What I said was: The net field, inside the sphere, due to all charges on and outside the sphere, is zero.

I didn't say anything about the effect of charges inside the sphere.
If the field inside the sphere due to all sources was zero, then Faraday cages would kill people.

Why is the net field at the center of the spherical shell not kq/a²?

 

[andrea96]

I don't. Please read more carefully:
What I said was: The net field, inside the sphere, due to all charges on and outside the sphere, is zero.

I didn't say anything about the effect of charges inside the sphere.
If the field inside the sphere due to all sources was zero, then Faraday cages would kill people.

Why is the net field at the center of the spherical shell not kq/a²?

Excuse me I haven't understood...
however kq/a^2 is just the field dues by the internal point charge, but the distribution on the inner surface dues a non-zero field because the shift of q from the center changes the distribution on the inner sufrace than the -q charge on the inner surface isn't simmetrically distribuired and dues a field inside the cavity. If you are agree with me that q1 and the charge on the external surface doesn't influence the system costituted of q and the distibution on the internal surface, I'll post the complete proceedings to deduce the result that I've written above... what do you think about it?

 

[Simon Bridge]

You mean like this:
http://www.df.unipi.it/~macchi/TEACHING/FISICA2/PROBLEMS/shellcharge.pdf

The problem is basically an exercise in method of images... you just have to separate the two geometries.

 

[andrea96]

You mean like this:
http://www.df.unipi.it/~macchi/TEACHING/FISICA2/PROBLEMS/shellcharge.pdf

The problem is basically an exercise in method of images... you just have to separate the two geometries.

If in the problem you have just posted, you disconnect the spher and you add a point charge q1 outdide, the field inside the cavity don't change... are you agree?