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Quick electric field question

by Crazymechanic
Tags: electric, field
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Crazymechanic
#1
Sep3-13, 03:09 PM
P: 853
Okay We know that a spherical or any other geometric form conducting uniform object has it's electric field on the outside of the object as that is where all the charge resides.

Now if we would have a sphere made of some insulating material with lots and lots of tiny little spherical balls in the material each of them being separated by some very small distance from the other , now each of the little balls would be connected with a small wire ad the wires arranges so that they don't add up to form a mesh and hence forming a faraday cage which would prevent the charge from being inside.
So because the little balls have their charge uniformly distributed around them that would mean that for this larger spherical object made of these balls the field would be also inside or not?
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Jolb
#2
Sep3-13, 03:57 PM
P: 419
Well, first remember that if you have a very crappy Faraday cage--for example, just some thin wires forming the edges of a cube--then the field will penetrate somewhat into the Faraday cage. If you had a spherical conducting mesh, a mesh with bigger holes would allow more of the field to penetrate inside, and smaller holes would allow less, and eventually you approach a conducting sphere which does eliminate all the field inside it--a perfect faraday cage.

Now let's imagine that instead of a boring mesh, we instead have your example: a conducting ball at each intersection point on the spherical mesh, but rather than being connected to nearest neighbors, each ball instead is connected by a wire to random balls around the mesh. As long as all the balls are connected through some network of connections, the effect is the same as if they were connected by nearest neighbors. It doesn't matter what strange path the charges must take from one ball to another--as long as all the balls are connected, the charges will move between the balls to cancel out any field inside the mesh (remembering that the bigger holes in the mesh, the more penetration, and the smaller the holes, the more perfect a faraday cage). It will NOT be the case that each little ball gets a uniform surface charge distribution--this is because each ball's surface charge interacts with the others'.

Keep in mind that if we have two charged spherical conductors near each other, their surface charge distributions will NOT be uniform--you could think that the charge on one conductor pushes away (or pulls) the charges on the other conductor, or alternatively you might think that each conductor gets an "image charge" from the other conductor's charge.

Indeed, a practical example of a faraday cage--say, a hollow sphere of copper [or maybe graphene if you want a true 2D conductor]--actually does resemble a mesh of balls on the molecular level [the balls being the atoms and the mesh intersections being the lattice sites], and the connections between atoms are often more messy than nice connections between every neighbor.
Crazymechanic
#3
Sep4-13, 11:39 AM
P: 853
Ok for the sake of simplicity let's imagine that the balls are not connected to anything with wires , they just somehow magically all are at the same potential, all are separated by the same small amount of distance and are placed in a sphere which holds them but which is an insulator.
Now what happens ? still the field is only outside?

Well if the answer to this turns out to be just a slight yes (which I think will not happen) then I will tell you how I thought to connect them so that they don't end up being a faraday cage type of structure.

Nugatory
#4
Sep4-13, 12:10 PM
Sci Advisor
Thanks
P: 3,663
Quick electric field question

Quote Quote by Crazymechanic View Post
Ok for the sake of simplicity let's imagine that the balls are not connected to anything with wires , they just somehow magically all are at the same potential, all are separated by the same small amount of distance and are placed in a sphere which holds them but which is an insulator.
Now what happens ? still the field is only outside?
If they are all magically maintained at the same potential, then in the limit as the number of balls approaches infinity and the distance between them approaches zero the surface will more and more closely resemble the surface of a perfect conductor, with the field not penetrating into the interior.

Jolb has already pointed out that that's what going on with a real conductor composed of real atoms, as opposed to an ideal conductor which is just a volume of equal potential.
Crazymechanic
#5
Sep4-13, 01:44 PM
P: 853
Ok I think I understand this but why do you say that the balls approach infinity or that the distance between them decreases? In my example the distanced between them was fixed , all the balls are some distance apart but they are held in place by a isolating sphere which is made from such material so that it doesn't conduct like plastic for example.

So normally when I have a single ball it has a uniform field around it ion all directions , in this case when the balls are in a spherical fashion but still don't touch do they fields interact wth one another in such a way as to make it a faraday cage and prevent the field from being to the inside of this structure?
mikeph
#6
Sep4-13, 01:54 PM
P: 1,212
you're describing a dielectric

each conducting mini-sphere will develop an electric dipole as the bound charges arrange themselves to oppose the electric field. the macroscopic result is an effective permittivity.

if the spheres are uniformly distributed then the permittivity will be uniform, I don't believe there will be any Faraday cage effect if the doped material (e.g. plastic) is perfectly insulating.
Crazymechanic
#7
Sep4-13, 04:00 PM
P: 853
Ok the thing here is quite simple.
One has a dielectric sphere , we all can imagine a plastic ball i guess can we.
Now in this plastic ball there are little metal spheres each say 1mm apart and if all of those metal spheres are at the same + potential and not connected by a wire mesh which would make a faraday cage sort of thing then we can assume that the field would be both outside and inside the plastic ball right?

Like with a wire , imagine a copper wire now make many turns and shape it like a sphere with the lower turns being smaller and the upper ones too and the ones in the middle wider. Now If you take only the one end of the wire and connect it to a high + potential it has it's field all around the wire both inside and outside of the wire wound sphere , as long as you don't take the other end of the wire and attach it to the same potential with an electric connection right?

I added a little crappy image just in case for a better visualization.
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Jolb
#8
Sep4-13, 05:01 PM
P: 419
Quote Quote by Crazymechanic View Post
Ok the thing here is quite simple.
One has a dielectric sphere , we all can imagine a plastic ball i guess can we.
Now in this plastic ball there are little metal spheres each say 1mm apart and if all of those metal spheres are at the same + potential and not connected by a wire mesh which would make a faraday cage sort of thing then we can assume that the field would be both outside and inside the plastic ball right?
As mikeph explained, if you're considering a material that's composed of a lot of small conducting balls that are insulated from one another (they don't exchange charge between one another), then that material will behave like a polarizable dielectric. But in an varying external field, you can't keep such a material at all the same potential [without adding and subtracting charge from each ball "by hand"]--if you did then by definition you would be requiring it to be a conductor and thus charge must be able to flow between the balls.

However, a thin sphere made of a polarizable dielectric [like a plastic sphere] will not function as a faraday cage. If we put the same charge on each small conducting sphere [assuming no external field], it would be as if we attached a charge to each point on the dielectric sphere. This is the case you're describing--so yes an external field will penetrate inside the charged dielectric sphere.

Like with a wire , imagine a copper wire now make many turns and shape it like a sphere with the lower turns being smaller and the upper ones too and the ones in the middle wider. Now If you take only the one end of the wire and connect it to a high + potential it has it's field all around the wire both inside and outside of the wire wound sphere , as long as you don't take the other end of the wire and attach it to the same potential with an electric connection right?

I added a little crappy image just in case for a better visualization.
If it's made of a conducting wire, then yes that spherical coil will function as a faraday cage.

You ask about what happens if we put a charge on that coil. Think about what direction the field would point at some point inside the sphere--for example the center point. Each segment of wire contributes a field which is equal and opposite to the field generated by the wire segment at the opposite point on the spherical coil--so the field cancels at the center [just like for a charged spherical conductor--and just like for the spherical conductor, the field also cancels at all other points on the inside.]

It will not be the case that adding a charge to that coil will cause a uniform distribution of charge on the (cylindrical) surface of the wire--the charge accumulation on one part of the wire will be affected by the charges on the parts of the wire all around the spherical coil, and the net effect will be that most of the charge will go to the outer part of the wire's cylindrical surface, just as for a wire mesh sphere. As explained before, regardless of the path the electrons must take to find a configuration which best cancels out the field on the interior of the sphere, they will nonetheless find it and attenuate the interior field like a faraday cage.
mikeph
#9
Sep5-13, 01:43 AM
P: 1,212
Quote Quote by Crazymechanic View Post
Ok the thing here is quite simple.
One has a dielectric sphere , we all can imagine a plastic ball i guess can we.
Now in this plastic ball there are little metal spheres each say 1mm apart and if all of those metal spheres are at the same + potential and not connected by a wire mesh which would make a faraday cage sort of thing then we can assume that the field would be both outside and inside the plastic ball right?
No, for the exact reason I gave in my answer.

Quote Quote by Crazymechanic View Post
Like with a wire , imagine a copper wire now make many turns and shape it like a sphere with the lower turns being smaller and the upper ones too and the ones in the middle wider. Now If you take only the one end of the wire and connect it to a high + potential it has it's field all around the wire both inside and outside of the wire wound sphere , as long as you don't take the other end of the wire and attach it to the same potential with an electric connection right?

I added a little crappy image just in case for a better visualization.
You can't connect one part of something to a "+ potential" with no mention of where ground is. There is no field without a potential difference.
Crazymechanic
#10
Sep5-13, 04:30 AM
P: 853
@Jolb for the sake of the thought I said that the metal balls held in place by the insulating sphere just are charged no wires connecting them present.The way I could do that with wire etc comes later but is not important to understand the principle of work here.

Ok so I get the picture , both with the metal balls and plastic sphere or the wire wound in a spherical fashion in both cases most of the field would be on the outside of the structures due to the field that points inward canceling out ?

@mikeph : you said:""" You can't connect one part of something to a "+ potential" with no mention of where ground is. There is no field without a potential difference. """

Is that really so? a thunderstorm cloud is miles away from a ground potential yet it builds up charge and has a field around it.


Okay so If I have the same scenario with either my plastic sphere and balls or the spherical wire and I now put a small but negatively charged sphere inside the structure does that make the positive charge which normally tends to reside on the outer surface of the lager structure to come inside as the positive field lines normally go towards the negative?
As Jolb said normally the positive field lines cancel out inside the sphere but if there is a striong negative charge in the middle of it then now the field lines have something to run into.
Ok elaborate on this situation.
mikeph
#11
Sep5-13, 01:13 PM
P: 1,212
E is defined as -grad(V). If I tell you my laptop has a potential of 10 V, that doesn't help you at all.

You need a spatially-varying potential in order for grad(V) to be nonzero, and spatially varying means we need at least two values for V at two different places. A lone potential measurement is physically meaningless, as it can correspond to any electric field imaginable without contradiction.

In the case of a thunderstorm, a few miles is nothing when electromagnetic fields are concerned! The charge is real, the field is real, but the potential is only read when defined with respect to a ground.
mikeph
#12
Sep5-13, 01:20 PM
P: 1,212
The unconnected balls will not act as a Faraday cage. The electric field within each individual ball will be zero, but inside the large plastic ball (outside a metal one) it will just be reduced. You are honestly describing a dielectric, once you shrink the metal balls to the size of atoms, they become atomic dipoles, lining up in any standard dielectric material.


You've changed the question- are the metal balls +ve charged? That will create an electric field, and it won't cancel out. Put a Gauss' law closed surface around it and you have net +ve charge inside it, which means you have electric field pointing outwards.
Crazymechanic
#13
Sep5-13, 03:39 PM
P: 853
Ok I see , we get some misunderstanding because it's not always clear to others what I want to say, So yes I know that inside a conducting sphere (like each of those small balls on the large insulating sphere ) the field cancels out.
But my question originally was about the field inside the large plastic or whatever insulator type of sphere due to the small evenly spaced little balls all around that larger sphere.
The questions was would such arrangement make the electric field have a value both inside the large sphere and outside ?

That's why I want to have this larger sphere made of these little conducting ones while the larger sphere itself to be an insulator to get an electric field inside the larger one which would be impossible if I would just make the larger one out of metal or any other conducting material and then set it on some potential.
Crazymechanic
#14
Sep7-13, 03:24 PM
P: 853
So is my reasoning in the last post right ? :)
Jolb
#15
Sep7-13, 05:26 PM
P: 419
Yes, more or less. If we take an insulating sphere [a hollow sphere] and then decorate it with some small, isolated spots of conductor [which do not exchange charge between one another], then the decorated insulating sphere would not function as a faraday cage. If you're talking about a solid sphere rather than a hollow sphere, then the lattice of conducting spheres will act like a dielectric medium within the sphere--still definitely not a faraday cage.


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