In the circuit shown, the equivalent capacitance between the points A and B is

C

Rearranging the circuit, the points E and D are at the same potentital (by symmetry). Then the capacity between E and D can be removed.

$\frac{1}{C'}=\frac{1}{C}+\frac{1}{C}$

∴ C' = C/2

$\frac{C}{2}$ and $\frac{C}{2}$ are in parallel.

Hence $C_{eq}=\frac{C}{2}+\frac{C}{2}$ = C

∴ (D)