GSO001 Problem 10

First, consider the state at the end of [Operation 1]. After sufficient time, capacitor $C_1$C1​ is fully charged and its potential difference equals the EMF $V$V.

The charge $Q_1$Q1​ and electrostatic energy $U_1$U1​ stored in $C_1$C1​:

$$\begin{aligned} Q_1 &= C_1 V \\ U_1 &= \frac{1}{2} C_1 V^2 \end{aligned}$$Q1​U1​​=C1​V=21​C1​V2​

Next, [Operation 2] is performed. When $S_1$S1​ is opened and $S_2$S2​ is closed, charge redistributes until both capacitors have the same potential difference $V'$V′.

By charge conservation:

$$C_1 V' + C_2 V' = Q_1 = C_1 V \implies V' = \frac{C_1}{C_1 + C_2} V$$C1​V′+C2​V′=Q1​=C1​V⟹V′=C1​+C2​C1​​V

The total electrostatic energy $U_2$U2​ at the end of Operation 2:

$$U_2 = \frac{1}{2} (C_1 + C_2) V'^2 = \frac{1}{2} \frac{C_1^2}{C_1 + C_2} V^2$$U2​=21​(C1​+C2​)V′2=21​C1​+C2​C12​​V2

The Joule heat $J$J equals the decrease in electrostatic energy by energy conservation:

$$J = U_1 - U_2 = \frac{1}{2} C_1 V^2 - \frac{1}{2} \frac{C_1^2}{C_1 + C_2} V^2 = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} V^2$$J=U1​−U2​=21​C1​V2−21​C1​+C2​C12​​V2=21​C1​+C2​C1​C2​​V2

Substituting numerical values:

$$\begin{aligned} J &= \frac{1}{2} \times \frac{(4.0 \times 10^{-6}) \times (6.0 \times 10^{-6})}{(4.0 \times 10^{-6}) + (6.0 \times 10^{-6})} \times (120)^2 \\ &= \frac{1}{2} \times \frac{24.0 \times 10^{-12}}{10.0 \times 10^{-6}} \times 14400 \\ &= 1.2 \times 10^{-6} \times 14400 \\ &= 17280 \times 10^{-6} \ \text{J} \end{aligned}$$J​=21​×(4.0×10−6)+(6.0×10−6)(4.0×10−6)×(6.0×10−6)​×(120)2=21​×10.0×10−624.0×10−12​×14400=1.2×10−6×14400=17280×10−6 J​

Converting to $\mu\text{J}$μJ:

$$J = 17280 \ \mu\text{J}$$J=17280 μJ

Therefore, the answer is $17280$17280.

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[Alternative: Using the analogy with reduced mass]

The Joule heat lost when capacitors are connected has exactly the same mathematical structure as the kinetic energy lost in a perfectly inelastic collision of two bodies. Here, capacitance $C$C corresponds to mass $m$m and potential $V$V corresponds to velocity $v$v.

Just as the energy lost in a perfectly inelastic collision equals the relative kinetic energy $\frac{1}{2}\mu v_{\text{rel}}^2$21​μvrel2​ (where $\mu$μ is the reduced mass $\frac{m_1 m_2}{m_1 + m_2}$m1​+m2​m1​m2​​), the energy lost in the circuit equals $\frac{1}{2}C_{\text{series}}(\Delta V)^2$21​Cseries​(ΔV)2, where $C_{\text{series}} = \frac{C_1 C_2}{C_1 + C_2}$Cseries​=C1​+C2​C1​C2​​ and $\Delta V$ΔV is the potential difference before connection.

Before connection: $C_1$C1​ is at $120 \ \text{V}$120 V, $C_2$C2​ is at $0 \ \text{V}$0 V:

$$J = \frac{1}{2} \times \frac{(4.0 \times 10^{-6}) \times (6.0 \times 10^{-6})}{(4.0 + 6.0) \times 10^{-6}} \times (120 - 0)^2 = 1.2 \times 10^{-6} \times 14400 = 17280 \ \mu\text{J}$$J=21​×(4.0+6.0)×10−6(4.0×10−6)×(6.0×10−6)​×(120−0)2=1.2×10−6×14400=17280 μJ