C2 and C3 will each have 10v across their terminals (this means C3 is at 10v versus gnd and C2 is at 20V versus gnd). There are many ways to look at it. Logically there is no way the circuit could favor C2 over C3 or vice versa for one. Charging a cap requires current and that same current must flow through the second cap (Kirchoff's Current Law). Any time current flows through a cap it either charges or discharges it. In this case the same current flows so they both get the same charge.
For example, say the caps are only 100pF and you put a meter on C3. The capacitance of the connection and the load impedance of the meter, though very high, may quickly discharge C3 to 0v and cause C2 to charge to 10v. There are many ways to look at what just happened but for starters Kirchoff's Voltage Law says C2 + C3 = 20v at all times.
The internal leakage current as well as any board leakage can cause the same problem, especially over a significant period of time. It gets worse if Vc2=20V, Vc3=0v and you remove the 20V source and replace it with a load. Current will flow out of C2 but in that direction it will charge C3 backwards, which will kill it if it's polarized.
Thus it is potentially unreliable to try to increase the voltage rating of the cap (the most plausible reason for doing this) this way without using something to stabilize the center voltage like overwhelming the leakage with an external voltage divider. Also zener diodes can be used.