Capacitors, charge and energy: a question

[Nick]

Basically, you need to think of this in terms of conservation of energy (Joules)- you are mixing charge in Coulombs and energy in Joules and assume all the energy is transferred to the second capacitor (it isn't). The final energy will be the same as the initial stored energy in C1, i.e. 0.5 * C1 * sqr(V1) (#1) where V1 is the initial voltage.

When you connect C2 in parallel with C1, some energy is transferred to C2, and consequently, as the total energy (assuming a lossless system) remains constant, the voltage will be V2 across both capacitors, so:

0.5 * C1 * sqr(V1) = 0.5 * (C1 + C2) * sqr(V2) (#2)

Dropping the 0.5 and dividing both sides by (C1 + C2) and sqr(V1) we get

1 + C1/C2 = sqr(V2/V1), i.e. V2/V1 = sqroot(1 + C1/C2) (#3)

So, the ratio of the new voltage, V2, to the original voltage, V1, is given by the above equation (#3) and the energy stored remains constant.

For safety, we can now work backwards to check our result...

Using equation #3, V2 = V1*sqroot(1+C1/C2)

Substituting in #2 and using #1, 0.5 * (C1 + C2) * sqr(V1*sqroot(1 + C1/C2)) = 0.5 * C1 * sqr(V1)

...which if you cancel through you will find 1 = 1 !! i.e. energy gets conserved.

HTH