The question at hand is:
"Two capacitors of capacitances $C_1$ and $C_1$ have charge $Q_1$ and $Q_2$. How much energy, $\Delta w$, is dissipated when they are connected in parallel. Show explicitly that $\Delta w$ is non-negative."
I'm confused about what the physical situation is. I took the assumption that these capacitors were somehow pre-charged, and then connected to each other in parallel without a voltage source. However, I don't understand how this would work. If a circuit is composed of only 2 elements, I don't see how they could be in any arrangement but series. Nevertheless, I tried solving it that way:
Two capacitors in parallel have the same voltage drop. Charge will be redistributed to make it the same voltage for both. Let $Q_1'$ and $Q_2'$ be the charges on the capacitors after they are connected. Now, picture the equivalent capacitor
$C_{eq} = C_1 + C_2 =$ $\frac{Q_1' + Q_2'}{V_f}$
conservation of charge:
$Q_1' + Q_2'= Q_1 + Q_2$,
$C_{eq} =$ $\frac{Q_1 + Q_2}{V_f}$
$V_f = $$\frac{Q_1 + Q_2}{C_1 + C_2}$
The initial energy of the capactiors is:
$U_0 = $$\frac{Q_1^2}{2C_1}$$+\frac{Q_2^2}{2C_2}$
$U_f = $$\frac{1}{2}$$(C_1 + C_2)V_f^2 = $$\frac{(Q_1 + Q_2)^2}{2(C_1 + C_2)}$
$\Delta U$$ = $$\frac{(Q_1 + Q_2)^2}{2(C_1 + C_2)}$-$\frac{Q_1^2}{2C_1}$$-\frac{Q_2^2}{2C_2}$
However, I don't see how this is necessarily non-negative.
Best Answer
This is a nice question. There is an obvious reason why the energy will be dissipated. Thomson's theorem states that given any charged conducting surface, the charges will rearrange to form an equipotential because that will have the lowest electrostatic field energy. Then, on connecting the capacitors, you can see why energy will be shed(dissipated) right?. Because on connecting, the 2 capacitors behave as one conductor, and by thomsons theorem, the least energy state(equipotential) is obtained. Hence energy is lost. Saying this was to make you realise that your derivation was correct, except the last term, which should be the other way around. That is the energy dissipated will be $\Delta w$= $\frac{Q_1^2}{2C_1}$ + $\frac{Q_2^2}{2C_2}$ - $\frac{(Q_1 + Q_2)^2}{C_1 + C_2}$ Now by Titu's Lemma we have for all non-negative reals $a_i$ and $b_i$ The following inequality:
$\Sigma$ $\frac{a_i^2}{b_1}$ = $\frac{(\Sigma a_i)^2}{\Sigma b_i}$. Clearly, This shows you that $\Delta$w is positive.