A 2.00 [tex]\mu[/tex]F capacitor is initially connected to a 12 Volt battery. In a separate circuit, a 4.00 [tex]\mu[/tex]F capacitor is connected to a 30 Volt battery. The capacitors are then disconnected from their respective batteries and connected to each other (positive sides are connected together, and negative sides are connected together). What is the final charge on the 2[tex]\mu[/tex]F capacitor
Qtot = Q1 + Q2
.5(C1 V12) + .5(C2 V22) = .5(C1+C2)V2
The Attempt at a Solution
I'm not sure if you use conservation of charge or conservation of energy here. If I use conservation of charge, I get a common V of 24V and charge of 48[tex]\mu[/tex]C on the 2[tex]\mu[/tex]F capacitor.
If I use conservation of energy I get a common V of 25.5 V and a charge of 51[tex]\mu[/tex]C on the 2[tex]\mu[/tex]F capacitor.
The question has some words italicized which means they are probably key words, but I don't really get it. As always, any help is greatly appreciated!