Capacitor charged/full indicator
The charge on a cap is directly proportional to its voltage.  So yes, comparing the voltage to a threshold just below the "full" limit should get you the signal you want.

200 V is high, so you need divide it down to a managable level.  This is done with a resistor divider, which is a term you can look up and get lots of information on.  For example, with a resistor divider of 1 M&Omega; and 10 k&Omega;, the voltage is divided by 101.  That means 200 V in will result in just about 2 V out.  Use whatever reasonable voltage reference you have as one input to a comparator and the divider output as the other.  Adjust the divider so that you get a little more than the voltage reference when the input is 200 V.

Note that this will drain the 200 V somewhat.  In the example above, it will drain it by 200 &micro;A, which will dissipate 40 mW.  Use higher resistances to drain less current, but not so high that the output impedance gets so high that it picks up too much noise.  I'd say 1 M&Omega; to 10 M&Omega; for the top resistor is about right.