Another thing to note is, in a set of capacitors connected in series, the smallest capacitor always sees the highest voltage. In fact, the same formula used for a resistive potential divider, can be used for two capacitors in series, if reciprocal of each capacitor is taken first.
C1 = 47nF
C2 = 22nF
Since both capacitors are in nF and it's only the ratio between them that matters here, we can ignore the nano part, which stops us having to deal with big or small numbers.
Take the reciprocal of each capacitor:
Y1 = 1/C1 = 1/47 = 0.02128
Y2 = 1/C2 = 1/22 = 0.04545
Now use the usual formula for a resistive potential divider:
V(Output) = Y2*V1/(Y1+Y2)
V(Output) = 0.04545*12/(0.02128+0.04545) = 8.173V
https://en.wikipedia.org/wiki/Voltage_dividerThe only trouble with the above calculation, is it assumes each capacitor is perfect and doesn't leak any current. In real life capacitors have a leakage current, which be unpredictable and will start to dominate, once the capacitors have charged up. In the previous example, if C2 has a higher leakage current (lower equivalent parallel resistance) than C1, the voltage across C2 will settle to a lower value, than C1 after awhile. If there's no DC present, for example the two capacitors are across a purely AC voltage source, then the above formula will work. If the voltage rating of both of the capacitors is higher, than the supply voltage, and it's not being used as a voltage divider, then the unequal leakage current can be ignored..
If the voltage ratings of the capacitors is below the total supply voltage, then the uneven leakage current could be an issue. The solution is to add voltage sharing resistors across the capacitors. If the ratio of the resistors is the same as the reciprocal of capacitor values, then the steady state and initial voltages will match. In the above example, if a 2.2M resistor is connected across C1 and a 4M7 resistor across C2, then the steady state voltage, will be similar to the initial voltage when the capacitors are rapidly charged.
One short cut which is useful is, if the values of both of the capacitors are the same, then the total capacitance halved and any voltage sharing resistors used can be the same value.