Pressing the switch discharges the capacitor (C) which causes the voltage across capacitor (Vc) to fall from 10V to 0V rapidly. When Vc falls below Vninv (here its 5V set by pot at non inv terminal) the output of op amp turns high ie +Vcc.
Releasing the switch causes the capacitor to charge through the series resistor (R) which causes the voltage to rise. Once it rises above Vninv. The output turns low ie -VEE. The output remains high as long as the capacitor takes time to charge from 0V to 5V. The resistor R determines the charging rate of capacitor. Use the below given equation to set your time.
T=-RC ln(1- Vninv/Vcc)
In my circuit Vninv is 5V , Vcc=10V , therefore eq becomes
T=-RC ln(0.5)
ln is natural log with base 'e'
