How stable

1. Jul 27, 2007

pardesi

consider the plates ofa capacitor suppose u just have two capacitors connected in series then consider the two plates not connected to the battery that form a seprate system .one of them has $$+Q$$ and the other $$-Q$$
then how is that there is no flow of charges between them because one of them repells electron(considering the net effect of that and it's 'conjugate' plate) and other attracts (considering the net effect of this it's 'conjugate' plate)

2. Jul 27, 2007

mgb_phys

In a real capcitor there is a flow because the insulating material between the plates is obviously never a perfect insulator.

It's a common fault in electronics designs to produce a circuit which doesn;t work where the capcitors never fully charge because of the leakage current.

3. Jul 27, 2007

pardesi

but the wires between the plates of two capacitors is atleast theoritically perfectly conducting

4. Jul 29, 2007

pardesi

does someone have any idea....

5. Jul 30, 2007

Creator

For the same reason a neutral piece of paper is attracted to a positivily charged comb.

It happens because, even though the piece of paper is originally neutral, the positive charge on the comb induces a dipole field on the paper.

Likewise in your case, each outer plate (connected to the battery) "induces" opposite charge on the inner plates. In effect the formerly neutral inner plates (connected together) 'act as if' they are a large "induced" dipole.

The +Q and -Q on the inner plates are only there because of induction. They don't neutralize each other for the same reason an 'induced' dipole doesn't neutralize itself.
The charges in an E field arrange themselves in such a way to try to neutralize the externally applied FIELD.

Creator

Last edited: Jul 30, 2007
6. Jul 30, 2007

pardesi

but my question is about the field between +Q of one plate and -Q of other plate

7. Jul 30, 2007

meopemuk

If I understand correctly, this is what you have in mind:

.... +Q1 -Q2 +Q3 -Q4
------| |------| |-------

Absolute values of all charges are equal (Q1=Q2=Q3=Q4), if both capacitors are equal. They have different labels to help the discussion. You are asking, why charges -Q2 and +Q3 do not move toward each other. Right? Indeed, there is attraction between them, and the wire presents a convenient path.

The answer is that -Q2 is also attracted to +Q1, and +Q3 is also attracted to -Q4. So, in the equilibrium situation it is unlikely that Q2=Q3=0. The equilibrium values of charges Q1=Q2=Q3=Q4 can be found by knowing capacitance values and the applied voltage.

Eugene.

8. Jul 30, 2007

pardesi

no not that the capacitors move towards each other...i was talking about the intermediate electrons in the wire ...just near te negative charge they are pushed and juts near the positive they are attracted so how come the charging in this segment ever stop

9. Jul 30, 2007

meopemuk

I haven't implied that capacitor plates are moving. I was talking about moving charges.

There is no electric field in the wire connecting plates -Q2 and +Q3. The field created by charges -Q2 and +Q3 is exactly compensated by the field created by charges +Q1 and -Q4. Therefore, electrons in this wire are not moving.

I am not sure if you agree with my explanation, perhaps, I didn't understand your question. What is the charge distribution among all four plates, in your opinion?

Eugene.

10. Jul 31, 2007

pardesi

actually it is this fact that is worrying if the charges are so distributed that the field cancel out then the charging never stops end evn if they didn't it wouldn't (may be...)
i don't have any idea sbout charge distribution

11. Aug 1, 2007

meopemuk

I am not sure we are talking about the same thing. The equilibrium condition is exactly when forces acting on each charge in the system are balanced. In this situation, the electric field in the wire is zero, so electrons are not moving along the wire, and the charges on capacitors plates are not changing.

Eugene.