Question

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 : $({\epsilon }_0-$ permittivity of free space)

• A. Zero
• B. $\frac{-q^2}{4\pi {\epsilon }_{0}d}$
• C. $\frac{-q^2}{4\pi {\epsilon }_{0}d}\left(3-\frac{1}{\sqrt{2}}\right)$
• D. $\frac{-q^2}{4\pi {\epsilon }_{0}d}\left(6-\frac{1}{\sqrt{2}}\right)$

Answer 1

Answer: (A)

Correct answer is (a) Zero