How a constant current of 23 mA flows from the capacitor when the transisitor turned on?
When the inverter output goes to 0V, Q1 effectively disappears from the circuit (the E-B junction is reverse biased) so you can just analyze Q2. When it is high, Q2 is off and it effectively disappears, so the capacitor charges through R1 towards +5V. 

With the inverter output at 0V, the lower side of R4 is at -5V and the emitter is at about -0.6V if the transistor is active, so the emitter current (and the collector current assuming it's active) are about 4.3V/0.2K = 22mA. 

There's a bit of current flowing through the 22K resistor which changes with the voltage and some other non-ideal things going on so it's not going to be exactly constant, but there's a region where the current is more-or-less constant so there is a more-or-less constant slope downward of about 100mV/ns at the collector of Q2, from the initial charge of the capacitor (5V or so) down to perhaps -500mV, after which the transistor saturates.