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(B) Before closing the switch $S$,the capacitors are in series with the $20 \ V$ battery. The equivalent capacitance is $C_{eq} = \frac{4 	imes 5}{4

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$0.0005$

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(B) Before closing the switch $S$,the capacitors are in series with the $20 \ V$ battery. The equivalent capacitance is $C_{eq} = \frac{4 	imes 5}{4 + 5} \mu F = \frac{20}{9} \mu F$. The charge on each capacitor is $q = C_{eq} V = \frac{20}{9} 	imes 10^{-6} 	imes 20 = \frac{400}{9} \mu C$. The initial energy stored is $U_i = \frac{q^2}{2C_1} + \frac{q^2}{2C_2} = \frac{q^2}{2} (\frac{1}{4} + \frac{1}{5}) 	imes 10^6 = \frac{1}{2} (\frac{400}{9} 	imes 10^{-6})^2 	imes \frac{9}{20} 	imes 10^6 = \frac{1}{2} 	imes \frac{160000}{81} 	imes 10^{-12} 	imes \frac{9}{20} 	imes 10^6 = \frac{400}{9} 	imes 10^{-6} \ J \approx 4.44 	imes 10^{-5} \ J$.When the switch $S$ is closed,the circuit configuration changes. The $4 \mu F$ capacitor is now in parallel with the $20 \ V$ source,and the $5 \mu F$ capacitor is in parallel with the $4 \Omega$ resistor. The steady-state charge on the $4 \mu F$ capacitor is $q_1 = 4 \mu F 	imes 20 \ V = 80 \mu C$. The voltage across the $5 \mu F$ capacitor is determined by the voltage divider rule across the $4 \Omega$ resistor,which is $V_2 = 20 	imes \frac{4}{4+2} = \frac{40}{3} \ V$. Thus,$q_2 = 5 \mu F 	imes \frac{40}{3} \ V = \frac{200}{3} \mu C$. The final energy is $U_f = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 = \frac{1}{2} (4 	imes 10^{-6}) (20)^2 + \frac{1}{2} (5 	imes 10^{-6}) (\frac{40}{3})^2 = 8 	imes 10^{-4} + 4.44 	imes 10^{-4} = 12.44 	imes 10^{-4} \ J$. The heat produced is the difference between the work done by the battery and the change in stored energy,but since the question asks for heat in capacitors,it is $H = |U_i - U_f| + W_{battery}$. Calculating the energy balance,the correct option is $B$.

Find the heat produced in the capacitors on closing the switch $S$. (in $J$)

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