How does this configuration not create E field inside the conductor?

[harmyder]

Suppose an uncharged hollow metallic sphere has a non-centered charge inside. In this case, the charge distribution and field lines are like this

Wouldn't such an uneven charge distribution on the inside and outside of the conductor create a field inside the conductor?

 

[Ibix]

Charges can move inside the conductor, and if there is an E field inside it they will move until there isn't one. Once they've stopped moving you will find that the charge distribution generates (inside the conductor) the exact opposite of the field from any external sources, cancelling it out. So yes, the charge distribution generates a field inside the conductor, one which exactly cancels the external field.

 

[harmyder]

What bothers me is that the field lines created by charge distributions on the metal surfaces are so unlike those I expect from the point charge. Yeah, maybe it is just hard to visualize how they cancel out each other.

 

[Ibix]

What bothers me is that the field lines created by charge distributions on the metal surfaces are so unlike those I expect from the point charge.

The field lines drawn on the diagram are the lines of the total field, point charge plus conductor charges. Field lines are integral curves of the field, so they don't just add up the way the fields do and it's quite hard to see from the diagram what the separate fields would look like.

It's also worth noting that the curved lines inside the hollow don't look wildly different from the field lines of two nearby charges (which is roughly what they should look like).

Yeah, maybe it is just hard to visualize how they cancel out each other.

I don't think this diagram is the best tool for visualising this. What you want is a diagram filled with a grid of little arrows, with each one pointing in the direction the E field points at its location. Then you can draw two separate diagrams, one for the E field of the charge and one for the E field of the conductor charges. The arrows in a pair of such diagrams do add by vector addition.

 

[Orodruin]

which is roughly what they should look like

It is exactly what they should look like.

 

[sophiecentaur]

What you want is a diagram filled with a grid of little arrows, with each one pointing in the direction the E field points at its location.

It seems to me that the Shell theorem is what counts and that's what your proposed pattern of tiny compasses would look like. It's the Potential that would be the same (zero) everywhere and the figure in post #1 implies that it's not - because the 'spacing of the lines' is arbitrary ( chosen and drawn by someone to 'look right'). It would take no energy to move that charge to any point in the sphere and the 'lines' would follow that condition from 0V to 0V.

 

[Ibix]

It seems to me that the Shell theorem is what counts and that's what your proposed pattern of tiny compasses would look like. It's the Potential that would be the same (zero) everywhere and the figure in post #1 implies that it's not - because the 'spacing of the lines' is arbitrary ( chosen and drawn by someone to 'look right'). It would take no energy to move that charge to any point in the sphere and the 'lines' would follow that condition from 0V to 0V.

I don't think the potential inside is zero. With what's in the diagram, that's because the charge distribution on the interior side of the sphere is not uniform, although the charge distribution on the exterior is. Moving the point charge will redistribute the charge on the interior surface, which will require work.

The method of images allows you to replace the charge distribution on the interior with a second point charge outside the sphere in order to get the field in the hollow.

 

[sophiecentaur]

The method of images allows you to replace the charge distribution on the interior with a second point charge outside the sphere in order to get the field in the hollow.

The method of images, as shown in your link, works for a plane reflector. The inside of the sphere is not a plane. The resulting image of the point charge inside is not a point but is distributed over all space at infinity. Using the 'lines of force' model, for any elemental area on the inside surface will be balanced by an elemental area on the opposite end of a line through the point charge. The same number of field lines will pass through the elements. Inverse square law and size of area yield a constant so no net force per unit charge in any direction. That diagram is not plotted accurately and it gives a wrong impression.

Same thing with the gravitational potential in a hollow sphere. There is no 'weight force' inside.

 

[Ibix]

The method of images, as shown in your link, works for a plane reflector.

It also works for a spherical one, as shown in the link in my previous post, in the section headed "Reflection in a conducting sphere". It includes a diagram with off center internal charges (although it's a rather more complex scenario).

That diagram is not plotted accurately and it gives a wrong impression.

I don't think you are correct here.

Same thing with the gravitational potential in a hollow sphere. There is no 'weight force' inside.

That's different - there is no "gravitational conductor" there. I'm not even sure what such a thing would be - perhaps two light concentric shells with a massive and compressable fluid forming a thin spherical shell between them? That would certainly lead to a non-symmetric distribution of mass in the fluid, invalidating an analysis by the Shell Theorem.

 

[sophiecentaur]

It also works for a spherical one,

whoops! yes of course - a spherical concave mirror brings rays from an object near the centre to a fuzzy image off-centre. The focus gets worse for points near the surface. In terms of the charge and its virtual (opposite charge?) image, there would be attraction which would be strongest with the charge at the centre of the sphere. There has to be something wrong with this or the charge placed at the centre would be in a stable position.
What I do wrong boss?