Six charges $+q, -q, +q, -q, +q$ and $-q$ are fixed at the corners of a hexagon of side $d$ as shown in the figure. The work done in bringing a charge $q_0$ to the centre of the hexagon from infinity is: ($\varepsilon_0$ = permittivity of free space)

• A
  $0$
• B
  $\frac{-q^2}{4 \pi \varepsilon_0 d}$
• C
  $\frac{-q^2}{4 \pi \varepsilon_0 d} \left(3-\frac{1}{\sqrt{2}}\right)$
• D
  $\frac{-q^2}{4 \pi \varepsilon_0 d} \left(6-\frac{1}{\sqrt{2}}\right)$