It's a bit unclear what exactly you are asking. I suppose you have seen a antenna like a dipole and it looks at first glance like just two wires that aren't connected to anything, therefore the question is how can current flow and power be drawn? If so, it would help if you clarified that. Some antennas, like a loop or folded dipole, are exactly the opposite in that they appear to be dead shorts at first glance.
In any case (if my interpretation of your question is correct), what you are missing is that antenna is no longer a open or short at its intended frequency. Antennas are not lumped systems, which means that different parts will be at different phases of the signal at the same time. In fact, antennas exploit this to help produce the large voltages and current it takes to radiate significant power.
Often resonance is involved. Fill a bathtub partway with water. Now put your hand in the middle and move it back and forth only a short distance in the end to end direction. Once you find the right frequency, you will see that you can get a lot of water to slosh back and forth despite only a relatively small motion of your hand. Note that at the peaks, the water at one end of the tub is high and the other low, and nothing is flowing. In between the water is roughly level but is flowing strongly in the middle. Also note that it takes very little force from your hand to cause this and keep it going, but you have to be moving your hand at just the right frequency. A little faster or slower and it doesn't work anymore.
That was resonance, and is exactly what is happening in a dipole. In the case of a dipole, that water level becomes voltage and the flow rate of the water becomes current. The feedpoint in the middle of the dipole is where a little current of just the right frequency is fed in, which causes resonant sloshing of charges back and forth in the antenna. The voltages created at the ends of the antenna can therefore be a lot higher than anything you put in.
This sloshing of current back and forth makes high voltages at the ends. Together with the high current in the middle, the antenna disturbs the local E and B fields in such a way that power is lost from the antenna into those fields. That power eventually organizes itself into a self-propagating wave we call radio.
Since real power is lost, the antenna must appear to have a resistive component from the driving circuit point of view. A ideal antenna used at exactly the right frequency will appear to be purely resistive, meaning all the power dumped into it gets transmitted. This happens very close to the best resonant sloshing frequency for most antennas. That also explains why antennas often only work well for a narrow frequency range.
So getting back to your question, the problem is you are analyzing the antenna at DC, which is completely irrelevant. You have to analyze antennas at the frequencies they are intended to radiate at. At those frequencies, a lot of other stuff happens so that they don't look like opens or shorts as they do at DC.
These were common before the transistor and called "Phototubes". Every 16mm sound projector in the US schools had one to read the sound track on the film.
With higher voltage and multiple plates - one designed to emit electron when struck by light. The others are meant to kick out more electrons when they are hit by an energetic electron and you get a cascade. These are "Photomultipliers" and are still used extensively for photon counting and any very low light application like fluorescence spectroscopy or phosphorescence lifetime measurement.
There is a very fast and sensitive star tracking application that uses a spinning mask in the optical path to a photomultiplier. I have not seen one in decades even though they have a much faster response than CCD based tracking.
Hamamatsu is one maker of both devices in many forms.
Best Answer
Yes, and believe it or not there is a closed circuit. A simple monopole antenna uses ground as the return path - the incoming radio wave hits the antenna structure and a current circulates between monopole and ground and there will be an impedance too: -
The graph above shows what the electrical impedance of the monopole is and how it is dependant on the antenna length (height) and the wavelength of the radio wave. So, at about one quarter wave length the monopole looks purely resistive and that resistance is about 37 ohms (hard to see on the graph I understand). That's the impedance it presents to the rest of the circuit.
This means your radio wave is transformed into a signal with an output impedance of 37 ohms. But free space / air has an impedance too - it's \$120\pi\$ or about 377 ohms and this is due to the capacitance and inductance of free space i.e. the physical fundamental properties that dictate the speed of light.
So yes, there is a closed circuit.
Here's an example - if you wanted to tune into an AM broadcast at 1 MHz you could construct a quarter wave monopole but, that monopole would be 75 metres long and present an impedance of 37 ohms.
Or you could make a 15 metre long (0.05 wavelength) monopole that presents a capacitive impedance of about 1000 ohms (or 159 pF at 1 MHz). You would get more signal from the quarter wave antenna but, it would be really big and cumbersome then, to tune it you'd need a more complex circuit than the 15 metre antenna because that shorter antenna already looks like 159 pF and can directly connect to a coil to give good station selectivity. That's what the olde worlde crystal set users did.
Regarding your other question I have no idea what you mean so further information such as a circuit might be required.