Charge distribution (Gauss law)

jeff1evesque

Claim:
A charged sphere will have it's charges disperse (uniformly for symmetrical objects) on the outer surface.

Proof:
Consider a sphere charged to some amount q. If we take small gaussian surfaces within the sphere, it will have [tex]\vec{E} = 0[/tex] on the closed gaussian sphere [since it is an equipotential surface]. Therefore, the electric flux is zero, so the electric charge from the small gaussian surface will have zero electric charge enclosed [Note: Gauss law states- the electric flux through any closed surface is proportional to the enclosed electric charge.

Now if we take enough small gaussian surfaces to fill the entire sphere, we find analogously the charge enclosed from all little gaussian spheres is zero. Therefore, the charge enclosed in the larger sphere is zero. But it has been charged to charge q, so the only possibility is that charge resides on the surface.

Question:
Is the proof above reasonable? Does this same reasoning apply to a single charged plate? Will charge distribute along the plate-surface where it was charged, and leave the other side uncharged [I can see why conducting spheres is a good shield in the inside, but cannot see how solid conducting plates can be used as shields respectively]?

kuruman

Your answer is in your question. As long as the charged plate is thick enough to have another "side", the charge that you place on the plate will split itself equally over both sides. If the "plate" is only a few atoms thick, then the classical E&M description breaks down and you have to use quantum theory.

jeff1evesque

So charge will not simply be on one side of the plate? Because, plate conductors is often used in electronics to shield the devices from nearby electric fields. If both of sides of the plate becomes charged, then the device facing either sides will be effected, and contradicts the whole notion of shielding.

cepheid

I think the idea is that the presence of an applied electric field causes the charges within the conductor to distribute themselves along (all) out surfaces in such a way that an electric field is set up within the conductor -- one that has the effect of cancelling out the applied field. So the net electric field inside the conductor is zero. This is how the shielding is accomplished. None of the applied electric field is allowed to penetrate to the opposite side of the conductor. I believe this is the principle of operation behind a Faraday cage.

jeff1evesque

Yea, I was trying to relate Faraday's concept to the shielding of plates. That makes sense, so the applied electric reacts to the plate. I just wasn't sure if the behavior was the same, or specifically how the electric field cancels (still not really sure, but I'll just accept it).

kuruman

Conductors act as electrical shields because they are equipotential surfaces and remain such. A grounded conductor is at zero electric potential. If charges move around a grounded conductor, electrons from the Earth will flow in or out of the conductor to maintain the conductor at zero potential. If you wrap a conductor completely around the device you have full shielding. If you are using only a grounded plate, the shielding is not as perfect, but the plate acts as a shield in the sense that the electric potential in its immediate vicinity is near zero.

cepheid

Don't just accept it. The explanation is simple. Let's say we have a flat horizontal conducting plate, and an applied electric field above it, pointing downwards (towards the top surface of the plate). In a conductor, electrons are free to move. The applied field will tend to accelerate electrons upwards (in the opposite direction of the field). Hence we have a concentration of free electrons at the top surface of the plate and a concentration of positive charge (positive ion cores that are now devoid of their electrons) at the bottom surface of the plate. However, this "separation of charge" sets up its own electric field. If you have positive charge at the bottom surface and negative charge at the top surface, in what direction does the associated field point? The answer is that it points upwards (away from the positive charges on the bottom surface, and towards the negative charges on the top surface). The electric field due to the separated charges points in the opposite direction from the electric field that was applied to separate them in the first place. Hence, the two fields tend to cancel each other out. (Indeed, charges will continue to be accelerated until the two fields exactly cancel...the whole process regulates itself).

EDIT: This is really a case where a picture is worth a thousand words. I'd recommend drawing one.

jeff1evesque

Got it. Also the electric field opposite to charge particle (creating the applied electric field) will cancel.

Thanks,


JL

jeff1evesque

Sorry to do this, but could you explain how the fields cancel on the opposite side of the particle that created the applied electric field? To me it seems the fields created by the positive and negative charges from the plate cancel outside the plate. But the electric field from the source (charge that induced applied electric field) is still present. So at the opposite end of the applied charge exists the applied electric field, is that correct?

kuruman

Correct. The electric field lines stop at the side of the conductor that has the negative induced charges and restart at the other side that has the positive charges. Inside the conductor there is no electric field. Cepheid is right. A picture is worth a lot in this case. I'll see if I can find one.

kuruman

OK, here is a picture as an attachment. The blue area is a piece of the conducting slab. Note that electric field lines start at + charges and stop at - charges. The black field lines that are generated by charges outside the conductor are canceled by the red lines so that the net field inside the conductor is zero as it should.

Attached Thumbnails

 

cepheid

The wikipedia page for "Faraday Cage" has a great animation.

jeff1evesque

Cool, that's exactly how I pictured it. However, if the "black field lines" continue on the other side of the conducting slab, then is it still a shield? It seems if outside electric field exists on the opposite side of a conductor, it does not have the qualities of a shield. But most circuit boards have a solid conductor "to provide shielding"- which is similar to what I am trying to understand. By what we've found, the applied electric field will surely influence the circuitry behind the shield.

cepheid

I guess for a single flat slab, that would be true. But look at the Wikipedia animation for Faraday Cage that I mentioned. If the electronics are fully enclosed (the shielding goes all the way around like a cage), then the separation of charge occurs between the "near side" and "far side" of the enclosure, leaving the hollow space within it devoid of any electric field.