Energy dissipated when two charged capacitors are connected in parallel

Question

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.

Answer 1

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.

Answer 2

I will show that the expression $\Delta U$ is in fact negative, rather than positive, since the system dissapates energy, rather than spontaneously gaining it. That is, we must have $\Delta U = - \Delta w$.

Starting from your expression for $\Delta U$, $$\Delta U = \frac{(Q_1+Q_2)^2}{2(C_1+C_2)}-\frac{Q_1^2}{2 C_1} - \frac{Q_2^2}{2 C^2},$$ we can find a common denominator $$\Delta U =\frac{ (Q_1 + Q_2)^2 C_1 C_2 - (Q_1^2 C_2 + Q_2^2 C_1)(C_1 + C_2) }{(C_1 + C_2) C_1 C_2}$$ and expand: $$\Delta U = \frac{Q_1^2 C_1 C_2 + Q_2^2 C_1 C_2 + 2 Q_1 Q_2 C_1 C_2 - Q_1^2 C_1 C_2 - Q_1^2 C_2^2 - Q_2^2 C_1^2 - Q_2^2 C_1 C_2}{(C_1 + C_2) C_1 C_2}.$$ Cancelling the first two terms in the numerator with their negative counterparts, $$\begin{align} \Delta U &= \frac{2 Q_1 Q_2 C_1 C_2 - Q_1^2 C_2^2 - Q_2^2 C_1^2}{(C_1 + C_2) C_1 C_2}\\ &= \frac{-(Q_1 C_2 - Q_2 C_1)^2}{(C_1 + C_2) C_1 C_2}.\\ \end{align}$$

Our final expression for $\Delta U$ is clearly negative, as the numerator contains only a negative sign and a squared number, and the denominator has only positive capacitances.

Answer 3

It is relatively easy to show that the system loses energy (your $\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}$ is negative by making it just one fraction and showing that the numerator is negative), what is not quite so obvious is how the system does this.

What you have to realise is that the system does not reach the $Q'_1$ and $Q'_2$ state instantaneously.
The point is that the circuit of two capacitors connected in parallel also has inductance and resistance.

So in reaching the final state the charge or current in the circuit undergoes damped harmonic motion just like any other LCR circuit, the frequency of the oscillations depending on the capacitance, inductance and resistance in the circuit.

As the inductance is going to be low the frequency of there oscillations is going to be high and if the resistance is also low there is going to be a lot of damping and the move towards the final state will be accompanied by the dissipation of heat in the resistor.
The system might well be over damped and so no oscillation of charge occur.

If the resistance is low then the oscillations last for longer and energy is still lost as heat in the resistor but proportionately more energy is lost to the system as electromagnetic radiation due to the charges in the circuit being accelerated as any accelerating unbound charge will emit electromagnetic radiation.

Answer 4

loss of energy when 2 capacitors are connected in parallel( -ive terminal with
-ive terminal of capacitors and +ive terminal with +ive terminal of capacitor) let, C1 capacitor is charged up to V1 potential. C2 capacitor is charged up to V2 potential. Q=CV initial total charge on the capacitors= (C1*V1)+(C2*V2)

common potential V= total charge/total capacity,
V=((C1V1)+(C2V2))/C1+C2 energy stored, E=(CV^2)/2 final energy= ((C1+C2)*V^2)/2 initial energy= (C1V1^2-C2V2^2)/2

FINAL ENERGY-INITIAL ENERGY=-(C1*C2(V1+V2)^2)/2*(C1+C2)