Since there is a capacitor, isn't it an incomplete circuit? How does the current flow across the dielectric medium? If a bulb was connected to a cell alone and if there was a break in the wires of the circuit, the bulb wouldn't light up. How does it light up with a capacitor?
- Connect the battery to this circuit. At this point, the charge doesn't know that there is a hole in the circuit.
- Negative charge therefor flows away from the negative battery pole, since it is repelled by this same charge, as far away as it can along the attached wire - this means that the charge will pass through the light bulb, and the light bulb will light up. The negative charge ends at the left capacitor plate.
- On the right-hand-side electrons will be attracted to the positive pole, and will move from the wire and closer to the battery. Also, electrons on the right capacitor plate will be repelled by the negative charge building up on the left plate. Electrons move away from the right plate, corresponding to positive charge, building up on the right plate.
- The charge keeps building up (accumulating) on the plates. Soon though the large amount of negative charge on the left plate will repel any more electrons that are coming. Same for the positive plate. Then the electron flow will stop. Current stops, and the light dims and turns off.
At first, the charges do not "see" the hole, but after a while they do. So an uncharged capacitor acts like a wire, as was it a closed circuit, but a fully charged capacitor acts like a hole, as was it an open circuit.
The point here is that there will be light, but only for a short while. If you remove the battery and just connect the wires, then the stored charge on the plates will move back again, and the light bulb will light up again, until there again is no more charge flowing. Current flows (and the bulb will light up) while the capacitor is being charged and while it is being discharged. Only.