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 $\left(\varepsilon_0-\right.$ permittivity of free space):

• Correct answer
  A

  Zero
• 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

  $\frac{-q^2}{4 \pi \varepsilon_0 d}\left(6-\frac{1}{\sqrt{2}}\right)$

Hence, the correct answer is option: 'A'.