Voltage = 400v

I = C*dv/dt. dv/dt is rate of change of voltage in volts per second - which means voltage would drop 1/.000047 or about 2130 volts per second. It will be back to zero volts in about one-tenth of a second.

Capacitor can only discharge slowly by limiting current or adding capacitance.

Charged capacitors can deliver high current at high voltage and is hazardous.

EDIT: The energy stored in the cap can be computed as follows:

E=1/2CV^2 = 0.5 ×47⋅10−6F × (400V)2=3.8J

You could add a resistor in series with the motor, so to drop the voltage to a sane level and limit the current (thus increasing the discharge time), but then only a fraction of that energy gets delivered to the motor, rather than just dissipated in the resistor!

For comparison, a AA battery (let's say 1.5 V, 2.4Ah) stores as much as 13000 joules!

Gold capacitor

Experiment

__Adding resistor with various resistance for the 47uF @ 400 Volt capacitor to light up LED__

10 Ohm - Still fast release

100 Ohm - Still fast release

270 Ohm - Still fast release

1KOhm - Still fast release

20KOhm - Release a little slower

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