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(D) When $n$ capacitors,each of capacitance $C$,are connected in series,the equivalent capacitance $C_{eq}$ is given by the formula:$1/C_{eq} = 1/C_1

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$2 	imes 10^{-12}\,F$

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(D) When $n$ capacitors,each of capacitance $C$,are connected in series,the equivalent capacitance $C_{eq}$ is given by the formula:$1/C_{eq} = 1/C_1 + 1/C_2 + 1/C_3 + ... + 1/C_n$Since all capacitors have the same capacitance $C = 6\,pF = 6 	imes 10^{-12}\,F$ and $n = 3$,the formula simplifies to:$C_{eq} = C/n$Substituting the values:$C_{eq} = (6 	imes 10^{-12}\,F) / 3 = 2 	imes 10^{-12}\,F$Therefore,the total capacitance of the combination is $2 	imes 10^{-12}\,F$.

In the figure,three capacitors each of capacitance $6\,pF$ are connected in series. The total capacitance of the combination will be

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