Will a voltage applied to an open circuit induce a brief current?

Question

I am wondering if a voltage applied to an open circuit will induce current in the circuit briefly as equilibrium is reached, despite no completed circuit.

For Example

Let's say I take a loop of copper wire, and then cut it in two places. This will create two semicircles of copper wire. If I then take a battery and connect the two semicircles at one end, the open end of the wires will have opposite charges, creating an electric field.

When the voltage was applied, it moved electrons from the wire connected to the positive terminal to the wire connected to the negative terminal. In some very short time, the system will come to an equilibrium and a capacitor of sorts will be formed at the open end of the wire loop.

Will the charging of this crude capacitor induce a current in the wire for a very short time? Will that current essentially be some exponential like this?

$$ i(t)=I_{max}(1-e^{t}) $$

What would $I_{max}$ be equal to? Does $I_{max}=\infty$?

Answer 1

It will indeed induce a real current in both pieces of the wire, for a brief instant. And here's the kicker: imagine now that you straighten out the two wires so one pair of cut ends are close together and the other ends are far apart. Then feed the close cut ends with nothing but brief impulses, (+)-going on one wire and (-)-going on the other. You now have a radio antenna, capable of radiating away hundreds of watts of electromagnetic waves!

Note that despite the fact that the far ends of the wire are not physically connected together- and hence no complete current loop path exists between the (+) and (-) terminals- real current flows through the two pieces of wire and real voltages are propelling the currents.

All you have to do is find the right "impulse" frequency with which to "feed" this antenna. For a run of wire 10 meters long (5 meters each for the two pieces) the right frequency is 30 mHz- which is in the shortwave radio band.