question_answer

A hexagon of side 8 cm has a charge 4uC at each of its vertices. The potential at the centre of the hexagon is:

A) \[2.7\times {{10}^{6}}V\]                    

B) \[7.2\times {{10}^{11}}V\]

C) \[2.5\times {{10}^{12}}V\]                  

D) \[3.4\times {{10}^{4}}V\]

Correct Answer: A

Solution :

(a) \[2.7\times {{10}^{6}}V\] As shown in the figure, O is the centre of hexagon ABCDEF of each side 8 cm. As it is a regular hexagon OAB, OBC, etc. are equilateral triangles. \[\therefore \,\,\,\,OA=OB=OC=OD\] \[=OE=OF=8\,cm=8\times {{10}^{-2}}m\] So, the potential at O is             \[V=6\times \frac{q}{4\pi {{\varepsilon }_{0}}r}\]             \[=\frac{6\times 9\times {{10}^{9}}\times 4\times {{10}^{-6}}}{8\times {{10}^{-2}}}=2.7\times {{10}^{6}}V\]