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

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