Voltage after connecting two capacitors

[cashmoney805]

Homework Statement

A 2.00 \muF capacitor is initially connected to a 12 Volt battery. In a separate circuit, a 4.00 \muF 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\muF capacitor

Homework Equations

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\muC on the 2\muF capacitor.
If I use conservation of energy I get a common V of 25.5 V and a charge of 51\muC on the 2\muF 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!

 

[LowlyPion]

You have a conservation of charge.

Figure the total charge.

Figure the equivalent C.

Figure the new Voltage, from the equivalent C and total charge.

Figure the charge across the C at the new voltage calculated.

 

[cashmoney805]

You have a conservation of charge.

Figure the total charge.

Figure the equivalent C.

Figure the new Voltage, from the equivalent C and total charge.

Figure the charge across the C at the new voltage calculated.

Qtot = (C1*V1) + (C2*V2) = 1.44 *10^-4C
Ceq = 6\muF
Qtot/Ceq = V = 24V

Q = C1*V = 4.8 * 10^-5

I can do the math, but I don't understand it conceptually. When would you use conservation of energy?

 

[LowlyPion]

I can do the math, but I don't understand it conceptually. When would you use conservation of energy?

There are only so many charges once everything is charged. You can't create them. They got to go somewhere. So apply the conservation of mass - as the conservation of charges.

Can't think of an example for using conservation of energy for this kind of problem. Voltage sources are usually doing work to keep things charged, and current flowing and resistors are dissipating power.

 

[cashmoney805]

Hm. In my textbook there is a similar problem. One capacitor is charged by one source, and a second one is charged by a second source. They are disconnected from their batteries and then connected to each other. Find potential difference across the capacitors and charge on each capacitor.

The book solves that problem by using conservation of energy. Anyone see the difference?