Irodov 3.131: Charge Survives, a Third Is Kept – Amit & Deepali Sinha Foundation

Problem Irodov 3.131

A capacitor $C_1 = 1.0$ $\mu$ F, charged to a voltage $V = 300$ V, is connected in parallel with an uncharged capacitor $C_2 = 2.0$ $\mu$ F. Find the change in the electric energy of the system once equilibrium is reached, and explain the result.

Before connection: C_1, charged to V (drawn for V > 0), and the uncharged C_2; the terminals are then connected in parallel, top to top and bottom to bottom. — Before connection: $C_1$ , charged to $V$ (drawn for $V > 0$ ), and the uncharged $C_2$ ; the terminals are then connected in parallel, top to top and bottom to bottom.

Given

$C_1 = 1.0~\mu$ F
$V = 300$ V
$C_2 = 2.0~\mu$ F, uncharged

Find

the energy increment

\Delta W

, and why it has the value it has

Solution

Step 1The charge is shared, not lost

The two capacitors form an isolated system once joined, so the charge $C_1V$ that $C_1$ held is conserved and spreads over both until they reach a common voltage $V'$ :

\begin{align*} C_1V &= (C_1 + C_2)\,V' \\ V' &= \frac{C_1}{C_1 + C_2}\,V \tag{1}\end{align*}

After connection: both capacitors at the common voltage V', the charge C_1V now shared between them in proportion to their capacitances. — After connection: both capacitors at the common voltage $V'$ , the charge $C_1V$ now shared between them in proportion to their capacitances.

Step 2The energy before and after

Initially only $C_1$ stores energy; at the end the pair stores the same charge on the parallel capacitance $C_1 + C_2$ :

\begin{aligned} W_i &= \frac{1}{2}\,C_1V^2\\[2pt] W_f &= \frac{(C_1V)^2}{2(C_1 + C_2)} \end{aligned}

The increment is their difference:

\begin{aligned} \Delta W = W_f - W_i &= \frac{1}{2}\,C_1V^2\left(\frac{C_1}{C_1 + C_2} - 1\right)\\[2pt] &= -\frac{1}{2}\,C_1V^2\,\frac{C_2}{C_1 + C_2} \end{aligned}

Numerically, with $C_1C_2/(C_1 + C_2) = 2/3~\mu$ F:

\Delta W = -\frac{1}{2} \times \frac{2}{3}\times 10^{-6}~\text{F} \times (300~\text{V})^2

\Delta W = -\frac{C_1C_2V^2}{2(C_1 + C_2)} \approx -30~\text{mJ}

(2)

The stored energy falls from $45$ mJ to $15$ mJ: two thirds of it disappear from the field.

Step 3Where the energy goes

The voltage u across the joining loop decays exponentially to zero, with time constant RC_s, while the current flows. — The voltage $u$ across the joining loop decays exponentially to zero, with time constant $RC_s$ , while the current flows.

Let $q_1, q_2$ be the charges and $u = q_1/C_1 - q_2/C_2$ the voltage driving the current $I = u/R$ round the loop, and write $C_s$ for the series capacitance:

C_s = \frac{C_1C_2}{C_1 + C_2} \tag{3}

Since $I$ drains $C_1$ and fills $C_2$ , the voltage obeys a first-order decay, which starts from $V$ :

\begin{align*} \frac{du}{dt} &= -\frac{I}{C_1} - \frac{I}{C_2} = -\frac{u}{R\,C_s} \\ u(t) &= V e^{-t/RC_s} \tag{4}\end{align*}

The total heat is the integral of $u^2/R$ over the transient, using (4):

Q = \int_0^\infty \frac{u^2}{R}\,dt = \frac{V^2}{R}\int_0^\infty e^{-2t/RC_s}\,dt = \frac{V^2}{R}\cdot\frac{RC_s}{2}

Q = \frac{1}{2}\,C_sV^2 = -\Delta W

(5)

$R$ cancels: however small the resistance, the same heat is released, only faster. (With very small $R$ part of the energy also leaves as electromagnetic radiation, but the total loss is still $-\Delta W$ .)

Check

Limits of (2). As $C_2 \to 0$ the second capacitor takes no charge and $\Delta W \to 0$ . As $C_2 \to \infty$ the large capacitor swallows the charge at almost zero voltage ( $V' \to 0$ by (1)), and $\Delta W$ tends to $-C_1V^2/2$ : all the stored energy is lost.

Remark

This is the electrical twin of a perfectly inelastic collision. Charge plays momentum, capacitance plays mass, voltage plays velocity and $CV^2/2$ plays $mv^2/2$ : the common voltage (1) is the common velocity, and the loss $C_sV^2/2$ is the familiar $\mu v_{\mathrm{rel}}^2/2$ , with the series capacitance (3) as the reduced mass.