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(C) The two $4 \,\mu F$ capacitors are in series,so their equivalent capacitance is $C_1 = \frac{4 	imes 4}{4 + 4} = 2 \,\mu F$.This $C_1$ is in para

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$0.4$

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(C) The two $4 \,\mu F$ capacitors are in series,so their equivalent capacitance is $C_1 = \frac{4 	imes 4}{4 + 4} = 2 \,\mu F$.This $C_1$ is in parallel with the $2 \,\mu F$ capacitor.Therefore,the net capacitance $C_{	ext{net}} = C_1 + 2 \,\mu F = 2 \,\mu F + 2 \,\mu F = 4 \,\mu F$.The total charge supplied by the battery is $q = C_{	ext{net}} 	imes V = 4 \,\mu F 	imes 10 \,V = 40 \,\mu C$.The work done by the battery is $W = q 	imes V = 40 \,\mu C 	imes 10 \,V = 400 \,\mu J$.Since $1 \,mJ = 1000 \,\mu J$,then $400 \,\mu J = 0.4 \,mJ$.

If the initial charge on all the capacitors was zero,the work done by the battery in the circuit shown is ........... $mJ$.

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