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Q=CV and if you do the math the capacitor voltage changes about 63% over the time RC. In your case it would be the difference between the current voltage and the 1.7V LED voltage.
If you know how to calculate the charge rate, then you know how to calculate the discharge rate, as the equations are the same; they depend on the capacitance and the circuit resistance. In your circuit, the capacitor discharges through the combanation of the 515 and 1K resistors. Thus, one time constant ( the time it take for the capacitor to discharge to 63% of it's voltage ) will be 550*10exp-6 * 1515 = 833mS. The voltage at that time will be (1-.63)*V2. It looks to me like the light will fade out in just under 1 second.
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