1) If we charge a stacked capacitor like the one in the picture and

[pmmphrkq]

1) If we charge a stacked capacitor like the one in the picture and then remove the leads (red and blue) what will be the voltage between bottom and top plate? Is it going to be 5 times (the number of stages) the original voltage or what?

[PLAIN]http://img89.imageshack.us/img89/4742/capr.png

2) If we take parallel plate capacitor and roll it up will the capacitance change? Surface area and distance between the electrodes will remain the same, but will the total area now increase (thus increasing the capacitance) because opposite electrodes will be close to each other (from the other side)? Or maybe electric field only exists between two plates?

Thanks in advance.

 

[Termotanque]

1) Good question. Except for the plates on top and below, the ones in the middle can be "sliced" in two so as to have 2 metallic. This enables one to think the problem as composed of 9 capacitors in parallel, namely: the one between the topmost plate (charge +q), and the top slice (charge -q) of the second plate as a whole (total charge -2q). The bottom slice of the second plate has a charge of -q, which forms a capacitor with the top slice (charge +q) of the third plate (total charge +2q).

Continuing in this fashion, you see that at equilibrium, the charges on the plates are, in order, +q, -2q, +2q, -2q, ..., +2q, -q.

Now remove the voltage source and suppose that the plates are not moved. The charges are conserved, so they are the same as above. Now you want the difference of potential between the topmost plate and the bottommost plate.

By definition, the potential is the path integral of the electric field from plate 1 to 10. Supposing that the distances from plate to plate is uniform, you have that the difference of potential between plate 1 and 2 is something we will call V0, and is negative because we are going from +q to -q. Next we are moving inside the conductor 2, so this adds nothing to the potential. Next we have the integral between 2 and 3, but this is exactly V0, except for the sign! And so they cancel out. You can see that this happens between plates 1,2,3 and 3,4,5 and 5,6,7 and 7,8,9 but nothing cancels the potential between 9 and 10, and so the required potential is exactly the same as the one of a single capacitor, the original voltage.

2) No, it does not change. Here is an example: imagine a plane plate capacitor of area A: its capacitance is Ae_0/d. Suppose now that you cut it in two, and now have 4 plates, each one of area A/2, and arrange them as the diagram in 1). They are in parallel, and so the equivalent capacitance is 2*(Ae_0/2d) = Ae_0/d.

 

[sweet springs]

Let me tell my thought to your question

1) If we charge a stacked capacitor like the one in the picture and then remove the leads (red and blue) what will be the voltage between bottom and top plate? Is it going to be 5 times (the number of stages) the original voltage or what?

It remains to original voltage.
Voltage of all the blue ternminals remain. Voltage of all the red ternminals remain.

2) If we take parallel plate capacitor and roll it up will the capacitance change? Surface area and distance between the electrodes will remain the same, but will the total area now increase (thus increasing the capacitance) because opposite electrodes will be close to each other (from the other side)? Or maybe electric field only exists between two plates?

If thickness by rolling is same as distance of plates, space storing electric field energy doubles, front and back. So I think capacitance doubles by rolling.

Please criticize. Regards.

 

[Philip Wood]

(1) Yes, voltage stays the same. Charges can't move.

(2) Capacitance doubles (as long as plates don't touch and separation is the same 'front and back'. The area of facing plate surfaces is doubled.