I was trying to use constant I to charge one capacitor.
The charging process is based on the equation: I*t=CV.
If C=1pF, t=8us, I=10nA, so V=I*t/C=80mV. The schematic simulation is very closer to 80mV.
However, if changing the C=100fF, the expected V=8us*10nA/100fF=800mV. The simulation result is less than 800mV.
I have no idea what results in the voltage reduction.
Thank you.

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Is it related to capacitor leakage current issue?
I read a article talked about the leakage current I is proportional to the  charges on the capacitor plate. For smaller capacitance, we got higher charge, so the leakage current is bigger. As a result, the final charged voltage could be smaller than expected.
Is that reasonable?
Thank you.

Or, if you don't want to play with gmin (which could affect other parts of your circuit), you could put a very large resistor in yourself.  Or, some simulators have a "switch" that you can have short to ground for the time=0 solution and then open up for t > 0.