What happens to surface charges when conductors/dielectrics touch?

  • Context: Undergrad 
  • Thread starter Thread starter feynman1
  • Start date Start date
  • Tags Tags
    Charges Surface
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
18 replies · 2K views
feynman1
Messages
435
Reaction score
29
When 2 conductors/dielectrics touch, will surface charges move away from their original conductor/dielectric to the other?
 
Physics news on Phys.org
Yes, they meet and cancel charges.
 
anuttarasammyak said:
Yes, they meet and cancel charges.
How could any charge move in a perfect dielectric? You are implying that something happens differently, the instant when they 'touch', compared with when there's a small gap. The Capacitance that the conductor will 'see', starts with an air spaced capacitor in series with a dielectric space. As the gap decreases, the capacitance of the air space increases without limit until all that's there is the dielectric layer. Can any charges flow onto or off the dielectric surface? By definition, I don't think so.

I think that the clue to the perceived paradox here is that the conductor is a real one and 'all the charges do not occupy an infinitely small region on the surface (the row of +++ signs that we draw in diagrams). The molecular fields in a real dielectric will not be uniform at the surface either.
 
  • Like
Likes   Reactions: feynman1
sophiecentaur said:
How could any charge move in a perfect dielectric? You are implying that something happens differently, the instant when they 'touch', compared with when there's a small gap. The Capacitance that the conductor will 'see', starts with an air spaced capacitor in series with a dielectric space. As the gap decreases, the capacitance of the air space increases without limit until all that's there is the dielectric layer. Can any charges flow onto or off the dielectric surface? By definition, I don't think so.

I think that the clue to the perceived paradox here is that the conductor is a real one and 'all the charges do not occupy an infinitely small region on the surface (the row of +++ signs that we draw in diagrams). The molecular fields in a real dielectric will not be uniform at the surface either.
I too disagree that charges in dielectrics can move, as there's no free charge there. But what about conductors?
 
Of course, in a dielectric the charges move a bit, but they are all bound, i.e., they don't move very far but are hold back by the binding forces (also electromagnetic). That's how a dielectric gets polarized by an external electric field. At a positively charged plate you due to this shift of charges in the dielectric you have a thin layer of negative charges at the boundary of the dielectric adjacent to the positively charged plate, and that's why the net effect of an dielectric within a capacitor enhances its capacitance by a factor ##\epsilon_r## compared to the capacitance of a "vacuum filled" capacitor, i.e., to get the same voltage difference ##U## you need to bring more charge ##Q=CU## to the plate in a dielectric-filled capacitor than in a vacuum-filled one.
 
By 'I too disagree that charges in dielectrics can move', I mean if 2 dielectrics touch, charges shouldn't exchange from dielectric to dielectric? Agree?
 
If everyone agrees that 2 dielectrics won't exchange surface charges when touching, what happens to 2 conductors then?
 
Can anyone draw a brief graph to illustrate this electrochemical potential difference between the metals? Thanks in advance.
 
hutchphd said:
Are the metals (conductors) the same or different? The electrochemical potential seems to unnecessarily complicate the discussion, but certainly is necessary for different metals.
Different conductors.
 
Say you connect a wire to the charged plates of conductor, charges move through wire so short circuit takes place. I do not see much difference on what take place in touching and in short circuiting.

I am not sure at all your setting of touching dielectric e.g. with E field keep working, touching with same/different polarity ?

In the usual case of sandwiching dielectric with condenser plates, surface charges of dielectric are cancelled, more exactly over cancelled, by charges of condenser plates.
 
  • Like
Likes   Reactions: feynman1
anuttarasammyak said:
surface charges of dielectric are cancelled,
I'm not sure why you mean by "cancelled". For two ideal dielectrics the electrons from one side will not flow onto the other side. But, as I wrote before, it is not fruitful to try to reconcile ideal situations with reality. It's called a discontinuity, I think.
 
  • Like
Likes   Reactions: feynman1
sophiecentaur said:
I'm not sure why you mean by "cancelled".
Please find an attached figure to show "cancel" I mean in post #14. Dielectric is also insulator, touching condenser plates.

2020-06-09 17.16.33.jpg
 
  • Like
Likes   Reactions: vanhees71
anuttarasammyak said:
Please find attached figures to show "cancel" I mean in post #14.
That's fine as an intuitive picture of it but how is that 'cancellation' any different from the same +- patterns that you have drawn all over the dielectric? Wouldn't you expect the gradient of the potential 'within' each molecule to be the same as the gradient of the potential between them? If it were not, there would be movement of charge to make it so.

In terms of field, you have a situation with constant field through the dielectric and Zero field within the conductor. It's just a matter of how 'micro' you want to go in describing the transition. Also, a real situation will have gaps between the two surfaces with a few points of contact. In those gaps you would have Air? or even a Vacuum?. Where would your "cancellation" be there?
 
  • Like
Likes   Reactions: feynman1
It's of course not completely cancelled. Since here the battery is connected to the capacitor you have a given voltage difference across the capacitor. That voltage difference leads to a charge ##\pm Q## on the plates with $$Q=C U = C_{\text{vac}} \epsilon_{\text{rel}} U,$$
i.e., you need more charge on the plate to get that voltage difference in the steady state than if the capacitor is filled with vacuum (because ##\epsilon_{\text{rel}}>1##). This is due to this "cancellation of charges" by polarization of the dieelectric.