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In a self-biased JFET circuit, the gate voltage must be approximately zero so that the reverse voltage at the gate-to-source will be equal(but negative) to the voltage across the source resistance. This can be achieved by putting very large resistance at the gate so that no current will flow.
BUT, what would happen if I left the gate open so that the infinite resistance will block all current? 
If you leave the gate open on a JFET, you end up with a circuit that will react to all variations in the electrical fields around it.
This is an electroscope that makes use of an open gate JFET to detect static electricity.
It is very sensitive to variations in the electrical field.
For an N-channel JFET: I suppose it would not conduct if there was stored negative charge on the gate since it wouldn't be able to drain out (or drain out very slowly). If it had zero or positive charge it would conduct.
Positive charge on the gate would be able to escape into the channel and equalize resulting in equivalent to Vgs = 0 (or slightly V>0 due to the PN junction voltage drop so not all positive charge could drain from the gate)
While any negative charge would be trapped and thus be floating relative to the source causing Vgs < 0.
Very similar to a MOSFET gate capacitance trapping charge when you disconnect the gate.
