Capacitor with connections on inner surfaces

[sridhar10chitta]

A capacitor C is made from large disks with a very large R and the gap between the plates s is very small (s<<R). The connections to the plates are made
"inside" the capacitor (on the inner surfaces) at the center of the plates. The capacitor is hooked up to a battery and switch. The circuit is closed. What will happen ?

 

[Meir Achuz]

Current will flow and the capacitor will charge until the potential difference equals V of the battery. The wires being between the plates changes nothing.

 

[sridhar10chitta]

fields inside the capacitor

will not the e-fields that are stronger inside the capacitor prevent the capacitor from charging ? could you describe a charging mechanism scenario ?

 

[Meir Achuz]

The E field inside the capacitor does not enter the wires.

 

[sridhar10chitta]

Why does not the E field enter the wires ?

 

[ranger]

Consider an infinite parallel-plate cap.

Why does not the E field enter the wires ?

For a infinite plate parallel plate capacitor, the e-field outside the capacitor is zero. If you consider placing two such plates in parallel with each other, each having different charges. An electric field will form between the plates (a capacitor). But what about the e-field outside? In order for there to be zero e-field, some cancellation must occur. This is exactly what happens:
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Capacitors/ParallCap.html

However with a finite plate capacitor, this is not the case i.e. the e-field outside the cap is not zero. However, for regions close to the capacitor, the plates will look as if they extend to infinity. This is similar to the above situation where we have zero e-field outside. Just remember that the potential falls off to zero as we approach infinity on an infinite plate capacitor. Zero potential = no e-field outside the cap.

 

[sridhar10chitta]

The e-field "inside" the capacitor will always be (Q/A) / epsilono and nearly uniform all over inside in the gap.

 

[pallidin]

Since there are no significant quantum effects in this arrangement, the capacitor will act as a "shorted" capacitor, rendering it's usefullness void.

 

[sridhar10chitta]

The anwer provided by Clem is apparently right. The battery will create the necessary conditions for the driving e-field inside the wires to exceed that developing on the plates and the "Current will flow and the capacitor will charge until the potential difference equals V of the battery. The wires being between the plates changes nothing."

 

[davidrit]

>>> The e-field "inside" the capacitor will always be (Q/A) / epsilono and nearly uniform all over inside in the gap.

This is only true for an ideal parallel plate capacitor and I don't think it applies precisely to your situation. Once you place the leads between the plates the real capacitor takes on a new geometry and the capacitance and distribution of the electric field change accordingly. In other words the outer surface of the wires are part of the capacitor plates. As a practical matter if the leads are much smaller than the plates then the effect will be minimum and you can use the ideal parallel plate equation as an approximation.

 

[sridhar10chitta]

David you are right. But let us not forget the orignal question which was "what will happen" and the answer is that the "battery will create the necessary conditions for the driving e-field inside the wires to exceed that developing on the plates and the Current will flow and the capacitor will charge until the potential difference equals V of the battery. The wires being between the plates changes nothing."