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 ( 0 - permittivity of free space) [figure]
- -q24 0 d (6 - 1 2 )
- Zero
- -q24 0 d
- -q24 0 d (3 - 1 2 )
Answer
(2) Zero
Solution
In a regular hexagon of side d, each corner is at distance d from the centre.
Potential at the centre:
V = 14 0 d(q - q + q - q + q - q) = 0
W = q0(Vcentre - V ) = q0(0 - 0)
W = 0
Answer: (2)