To expand i simple terms, imagine the following:
A battery with two terminals + and -
A 1 ohm resistor.
Let say the battery is, to make things easy, 1volt (ie the chemisty i the battery pushes the +ve terminal to an electrical potential that is 1volt higher than it's negative terminal)
Now put the 1ohm resistor across the battery + and -ve terminals. Ohms law tells us that 1 amp will flow (V=IR) because one end of the resistor is at 1v, and the other at 0v
we also know that the resistor must disipate some energy, i this case, 1Watt (W= V X I)
If we now pick a battery with a higher potential difference, say 10V, we have 10v across our 1 Ohm resistor, 10Amps flow, and the resistor disipates 100W as heat
At some point, the resisitor burns out, because it can't disipate the heat without actually going overtemperature and melting!
However, there is another case, and this time, we simply connect one end of our 1 ohm resistor to the +ve battery terminal, and leave the other end unconnected, in free air.
Here, there is no currently flow, and both ends of the resistor have 1v on them. In fact of course, we still have a complete circuit, it's just that part of the circuit is made of air, and not copper. As air has a relatively high resistance, very little current flows through it. However, if we increase the battery voltage, at some point (in the thousands of volts range) even this air gap breaks down and current flows. Those mechanisms are more to do with other factors such as the ionisation threshold of the air, and shape of the conductors, and the impurities (such as dust, or water vapour) carried in the air.