Would electrons flow around the loop continuously in the following diagram If there is a plate that is negatively charged (like the negatively charged plate of a capacitor) and it is brought near to (but not touching) a copper wire loop with a light bulb (load) and a diode (so current can only flow in one direction) would current flow continuously/perpetually for a long time and thus power the light bulb while it does. My thinking is that the electrons in the wire will be repelled by the negatively charged plate and want to flow through the current- like a pump in an annulus (ring) shaped swimming pool.

 A: If you convert this to a pressure system it becomes quite obvious why it doesn't work. The negative charge becomes a high pressure tank. Here we have the correspondence high pressure = high energy = high voltage. The circuit becomes piping. The diode becomes a one way valve and the light bulb can be anything (take a turbine for example). The high pressure air wants to escape to the piping (wire) of the circuit because the piping has a lower potential. 
If you open the valve of the high pressure tank the air rushes in. It spins the turbine and moves around the circuit. But this can't go on forever: soon every part of the pipe will reach the same pressure and once that point is reached no air will flow anymore. The same will happen in your case. After a very short time period every point in the circuit will have the same potential and no more current will flow. The diode won't change that.
A: On a fundamental level, your circuit is just an object that will acquire an induced charge.  Electrons on the near side of the circuit will feel a force from the negative charge on the plate, flow away from the plate, and redistribute themselves on the loop of wire.  Eventually, enough electrons will accumulate at the far end of the object that an equilibrium will be reached:  an electron anywhere on the circuit will feel equal repulsive forces from the negative charge on the plate and from the accumulation of electrons elsewhere in the circuit, and so it will stay put.
During the redistribution of the charge in the circuit, a current will flow very briefly.  But it is not likely to be detectable, and will certainly not be long-lived.
