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(D) From the circuit diagram,we can see that the two capacitors in the middle branch are connected in series. Let their equivalent capacitance be $C_s

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$0.1$

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(D) From the circuit diagram,we can see that the two capacitors in the middle branch are connected in series. Let their equivalent capacitance be $C_s$. Since they are identical,$C_s = \frac{C_1 	imes C_1}{C_1 + C_1} = \frac{C_1}{2}$.This equivalent capacitor $C_s$ is in parallel with the top capacitor $C_1$ and the bottom capacitor $C_1$.Therefore,the total equivalent capacitance $C_{eq}$ between points $A$ and $B$ is:$C_{eq} = C_1 + C_s + C_1 = C_1 + \frac{C_1}{2} + C_1 = \frac{5}{2} C_1$.Given that the battery voltage $V = 6 \, V$ and the total charge $Q = 1.5 \, \mu C$,we use the formula $Q = C_{eq} V$:$1.5 	imes 10^{-6} \, C = (\frac{5}{2} C_1) 	imes 6 \, V$.$1.5 	imes 10^{-6} = 15 \, C_1$.$C_1 = \frac{1.5 	imes 10^{-6}}{15} = 0.1 	imes 10^{-6} \, F = 0.1 \, \mu F$.Thus,the correct option is $D$.

Four identical capacitors are connected as shown in the diagram. When a battery of $6 \, V$ is connected between $A$ and $B$,the total charge stored is found to be $1.5 \, \mu C$. The value of $C_1$ is ........ $\mu F$.

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