Physics - Electrostatics Question with Solution | TestHub

As shown in the figure, $O$ is the centre of hexagon $ABCDEF$ of each side $8\mathrm{cm}$. As it is a regular hexagon $OAB,OBC$, etc are equilateral triangles. $$\begin{array}{rl}\therefore OA=OB=OC=& OD=OE=OF=8\mathrm{cm}\\ & =8\times 10^{-2}\mathrm{m}\end{array}$$ The potential at $O$ is $$\begin{array}{rl}V& =6\times \frac{q}{4\pi {\epsilon }_{0}r}\\ & =\frac{6\times 9\times 10^9\times 4\times 10^{-6}}{8\times 10^{-2}}=2.7\times 10^6\mathrm{V}\end{array}$$