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 q₀ to the centre of the hexagon from infinity is (ε₀ permittivity of free space)

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zero

$\frac{- q^{2}}{4 {πε}_{0} d}$

$\frac{- q^{2}}{4 {πε}_{0} d} (6 - \frac{1}{\sqrt{2}})$

$\frac{- q^{2}}{4 {πε}_{0} d} (3 - \frac{1}{\sqrt{2}})$

answer is A.

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Detailed Solution

Work to bring a charge ' q ' from infinity to a point is W=qV.
Where V is the potential at that point.
Here, potential at the centre of the hexagon is due to the charges present at its corner.
Question Image
$V = \frac{3 q}{4 {πϵ}_{0} d} - \frac{3 q}{4 {πϵ}_{0} d}$
where ' d ' is the distance of the corner to centre (as six equilateral triangles are formed when we joined the corners to centre in a hezagon).
So, V=0 and W=0
So, no work is done to bring a charge from infinity to the centre of the given hexagon.