Talk:Faraday's ice pail experiment

Latest comment: 4 years ago by Chetvorno in topic 3rd explination

2nd explination edit

I think the explanation should be repeated "in other words" to give new students a better change of getting a grasp of the situation.


This is another explanation that may be simpler.

For a charged rod and uncharged sphere: only when a rubber charged rod (+) influences an un-charged metal sphere does a (conducting sphere) develop opposing charges on each half of sphere (negative charges migrate toward positive of rubber positive leaving the other side of the sphere +, a simple matter of distance and balancing of opposite forces, the electrical pressure is balanced within the mobility constrains of the conducting sphere boundaries). Without any grounding when the rod is taken away, the sphere is again un-charged (it remains uncharged always - but a voltage appears between sides temporarily when the rod is introduced). Put another way: if a "+ charged conductor" were to gain electrons to balance the + charge on the outer sphere with a layer of - charges in more inner layer: the sphere would no longer be a charged sphere; it would be a un-charged sphere because all charges would be opposed and equalled.

For a charged sphere: E is zero but point voltage is not. The force between like net charge evenly distribute along the outer conductor because it is the easiest path. The e are restricted to the conductor at low charge/voltage to that of air - and to be dispersed that farthest from each other they obviously choose the surface). It is a simple matter of all pushing away from one another (like charges repel) and the medium and size they have to work within (the e fit on the surface without running out of space, so to speak). E==0 however the electrical potential (Vp) is not zero inside, it is the same as that near the outside of the sphere of course. Anywhere inside (not including center) there is an equal an opposite reaction for distance and number of pushes from that distance: simply put - if all forces are equidistant in a sphere (of any kind, electrical or otherwise), anywhere inside that sphere has an equal push from all sides (meaning E, the net direction of pull, is 0, no direction pulls more than another).

The last two paragraphs are FAR DIFFERENT than the purely mathematical [Gauss's law] which shows net flux "through a conductor" is zero. That is a simple matter that opposing fluxes are not counted and sphere has equal area seen from from both sides.

Induction in this situation (only): the charge on the inserted sphere repels the point voltage of the (bucket, outer sphere) on approach to the outer sphere (but is forced to stay within the conductor it cannot leap out, and are forced closer and inside). Upon arrival the inserted sphere has gained potential energy equal to that of of the bucket (work has been done, W=-U). Because E=0 (point voltage same everywhere) and the INSIDE of the outer conductor has zero charge (all charge is on outside of bucket, surface charge), the inside of the sphere is "an un-charged sphere" and any new electrical pressure must be dispersed equally to it as would any uncharged sphere. Before touching the effect is like the rubber rod and uncharged sphere above. The only force inside the outer conductor (where E=0 everywhere) is the charge (excess electrons) on the inserted sphere - and the outer un-charged sphere (the inner layer of the outer sphere) will equal to it. Next, the transferred electrons quickly redistribute to the outer shell and the outer shell now has higher Vp but inside is E=0 again: the process repeats until the inserted sphere is (+- 1e) un-charged. The whole thing can be compared to reason hydraulics work, very roughly speaking. But the situation is very specific to the fact electrons are "restricted to a conductor and it's shape" - it does not show allot of generality it is more of a paradox to explain.

Re your last paragraph, before the inserted sphere is touched to it, the inside of the bucket is not uncharged, it has a surface charge which is equal and opposite to the charge on the outside of the bucket, since the bucket started out with zero charge. This can be confirmed experimentally by inserting the electrometer probe into the bucket and touching it to the inside surface. --ChetvornoTALK 19:50, 5 August 2019 (UTC)Reply

3rd explination edit

A third explination or take-off from the 2nd is this.

Charges on the sphere maintain their electrical pressure but equally distribute - and because of that at any point inside sphere (inverse square fall-off and the count of charges and distances to them for a sphere shape) the pressure inside outer shell is equal at all points (but not zero). The inserted charge (it was forced inside the sphere, say 1e) maintains it's own electrical pressure (exactly 1e) at all times and the outer sphere is effected by it. The inserted charges "proper place" is equidistant from all other charges and if connected by conductor it will go there to it's place of equal distances (to the outer shell), because like charges repel. The argument is different if the "inserted charge" is of opposite charge sign, of course, the effects different too.

With the above understanding one may present that "electrical induction, applied to faraday's ice pail, is a fictional force" akin to Centrifugal force.

Neither the bucket nor the charged object need to be a sphere - they can be any shape. --ChetvornoTALK 01:17, 6 August 2019 (UTC) Also, you can sign your posts by typing ~~~~ after them. --ChetvornoTALK 01:17, 6 August 2019 (UTC)Reply