Question: 39: 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)
(1) Zero
(2) \frac{-q^{2}}{4\pi\varepsilon_{0}d}
(3) \frac{-q^{2}}{4\pi\varepsilon_{0}d}\left(3-\frac{1}{\sqrt{2}}\right)
(4) \frac{-q^{2}}{4\pi\varepsilon_{0}d}\left(6-\frac{1}{\sqrt{2}}\right)
Answer: Option (1)
Explanation:
The work done in bringing a charge from infinity to a point in an electrostatic field is equal to the electric potential at that point multiplied by the charge,
i.e. W=q_{0}V.
The electric potential at the centre of the hexagon is the algebraic sum of potentials due to all six charges.
The distance of each corner charge from the centre of a regular hexagon of side
d is equal to d.
The potential due to a charge q at distance d is
\frac{1}{4\pi\varepsilon_{0}}\frac{q}{d},
and due to -q is \frac{1}{4\pi\varepsilon_{0}}\frac{-q}{d}.
Since the charges are placed alternately as +q and -q at equal distances from the centre, the potentials due to opposite charges cancel each other pairwise.
Thus, the net electric potential at the centre of the hexagon is V=0.
Therefore, the work done in bringing the charge q_{0}
from infinity to the centre is W=q_{0}\times0=0.
Hence, the correct option is (1).