Capacitors and energy help plz!?

Question

A 2.21 µF capacitor is charged by a 20.0 V battery. It is disconnected from the battery and then connected to an uncharged 3.60 µF capacitor
http://www.webassign.net/giancoli5/17_28.gif
Determine the total stored energy after they are connected

Answer 1

As the capacitor is charged, the charge stored in the capacitor is q = CV = 2.21 E-6 × 20 = 44.2 E-6 C. Putting the second capacitor in parallel increases capacity to 2.21 µF + 3.6 µF = 5.81 µF. The energy stored in the capacitor is given by

U = ½ q²/ C = q² / 2C = (44.2 E-6)² / 11.62 E-6 = 168 E-6 = 168 µJ

This problem, or rather, some variations of it, have been around for a fairly long time. One of these variations I had to solve while I attended College. I recall that we were asked to find the energy stored in the first capacitor, and then, the energy stored in the compound capacitor; furthermore, we had to explain where the missing energy went to.

Because there was, actually, a loss of energy. In the present case, initial energy stored amounts to

U₀ = (44.2 E-6)² / 4.42 E-6 = 442 E-6 = 442 µJ

It was generally agreed that the solution had to conform to the Law of Conservation of Charge; in addition, voltage had to be the same in both capacitors, since they are connected in parallel.

Energy, by the way, is definitely NOT conserved in the process. If you repeat the experiment, you'll notice that a rather hefty spark jumps as capacitors are connected in parallel. Unquestionably, the spark represents a sizable fraction of the energy initially stored at the capacitor.

In their book, Mssrs. Halliday and Resnick sustain the opinion that energy is lost in the wire used to connect both capacitors in parallel. I doubt this, because the amount of energy lost seems to be totally independent of the means employed to make the connection; i.e., the same energy will be lost whether you use a thin, long wire or a thick, short one. For me, missing energy is taken up by the spark. Discussing this with my Physics teacher, he only remarked that it was known that charge, when subject to acceleration, radiates energy.

Pick your choice.

Answer 2

For ideal capacitors the total stored energy would not change. So since the initial energy is (CV^2)/2 then the energy would be

(2.21 * 400)/2 = 442 joules this will not change even when another capacitor is connected.

Answer 3

E=0.5cv*v
=0.5*2.21*20*20/1000000
=4.4/10000 (J)