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(D) From the circuit diagram,we can identify the arrangement of capacitors:$1$. The two $ 100 pF $ capacitors are connected in series. Their equivalen

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$ \frac{100}{3} pF $

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(D) From the circuit diagram,we can identify the arrangement of capacitors:$1$. The two $ 100 pF $ capacitors are connected in series. Their equivalent capacitance $ C_1 $ is given by: $\frac{1}{C_1} = \frac{1}{100} + \frac{1}{100} = \frac{2}{100} \implies C_1 = 50 pF$.$2$. This $ C_1 = 50 pF $ is in parallel with the top $ 50 pF $ capacitor. Let this parallel combination be $ C_2 $. $ C_2 = 50 pF + 50 pF = 100 pF$.$3$. Finally,this $ C_2 = 100 pF $ is in series with the bottom $ 50 pF $ capacitor connected to terminal $ B $. The total equivalent capacitance $ C_{AB} $ is: $\frac{1}{C_{AB}} = \frac{1}{100} + \frac{1}{50} = \frac{1+2}{100} = \frac{3}{100}$.$4$. Therefore,$ C_{AB} = \frac{100}{3} pF $.

The equivalent capacitance between $ A $ and $ B $ is,

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