Solving capacitor voltage "drops"?
"The charge on C1 and C2 must be equal" doesn't explain it. Why must they be equal?  

Just think of it as a voltage divider, but while for resistors \$  V_{R2} = \dfrac{R2}{R1 + R2} V1 \$, for capacitors: \$  V_{C2} = \dfrac{C1}{C1 + C2} V1 \$. Notice the different indices in the numerator.   



Proof  


  WARNING. Ugly equations ahead. Don't worry if they look too complicated; all you need in practice is the equation above.  


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\$ V_{C2} = \dfrac{Z_{C2}}{Z_{C1} + Z_{C2}}V1 = \dfrac{\dfrac{1}{j\omega C2}}{\dfrac{1}{j\omega C1} + \dfrac{1}{j\omega C2}} V1 = \dfrac{\dfrac{C1}{j\omega C1 C2}}{\dfrac{C2}{j\omega C1 C2} + \dfrac{C1}{j\omega C1 C2}} V1 = \dfrac{C1}{C1 + C2} V1 \$  

The last step is only allowed for \$\omega \ne 0\$. For DC we have to take the limit:  

\$  V_{C2} = \displaystyle \lim_{\omega \to 0} \dfrac{\dfrac{C1}{j\omega C1 C2}}{\dfrac{C2}{j\omega C1 C2} + \dfrac{C1}{j\omega C1 C2}} V1 = \dfrac{C1}{C1 + C2} V1 \$  

You don't need to do this derivation each time, just remember that it's the inverse of a resistor divider.