Low K helps by reducing the intra-metal capacitance, and propagation delay of signals. The material needs to be used with care as there are multiple processing issues that need to be overcome to make it reliable. The maximum use able metal to metal dielectric field needs to be considered, as the maximum field tolerated by low-K is generally lower then oxides, ( unless the oxide is very poor..)
The question should not be "why can...", but "why are low-k materials used" (not exactly for interconnects but to isolate the different metal interconnect layers).
As Matthew already stated, everything is about reducing capacity.
Inter-layer capacitance has 2 effects:
1. When switching, the charge stored in these capacities has to be changed - resulting in switching losses.
2. The propagation speed of the signals depends on line inductance and capacity. Line inductance depends mainly on line geometry thus cannot be changed for a given geometry. Line capacity depends on the 'k' of the insulation layers and layer thickness (also considering geometry as 'given').
The lower k, the lower the capacity - the higher the propagation speed == the lower the delays.
The interconnects are printed metal paths connecting between two terminals of devices and circuits. Their function is to transmit the signal and the power between the elements and the component of the circuit. In advanced very large scale integrated circuits, these interconnects play an appreciable role in the switching speed of the circuit and the highest operating frequency. Unfortunately, as the transistors are scaled down the interconnects size must also be reduced which increase the the interconnect resistance. For the same wire length its resistance even dominates the propagation speed. The signal delay of the interconnect is determined by its resistance and its capacitance with the return wire. So, in order to reduce the delay time, one reduces the inter layer capacitance by using low dielectric constant material. This has been explained by the colleagues Matthew and Dreher and i do not want to repeat their answer..
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