It is easiest to consider problems like this by thinking exclusively about parallel plate capacitors for which [tex]C \equiv \frac{Q}{V} =\kappa \epsilon_0 \frac{A}{d}[/tex] where Q is the charge separated (+Q on one plate, -Q on the other), V is the voltage difference between the plates, A is the area of each plate, and d is the separation between the plates.
When capacitors are connected in parallel, the voltage across each capacitor is the same. But with two capacitors, it will require more charge to reach the voltage V than it would with just one capacitor. In fact, if capacitor 1 requires charge
