Question

A regular hexagon of 10 cm side has charges of 5 u C at each of its vertices. calculate the potential at its centre.

Answer 1

Distance, $r=10\times {10}^{-2}m$

Charge, $q=5\times {10}^{-6}C$

Potential at point O is

$V_O=V_A+V_B+V_C+V_D+V_E+V_F$

$V_O=\frac{1}{4\pi {\epsilon }_0}[\frac{q}{OA}+\frac{q}{OB}+\frac{q}{OC}+\frac{q}{OD}+\frac{q}{OE}+\frac{q}{OF}]$

$V_O=9\times {10}^9[6\times \frac{5\times {10}^{-6}}{10\times {10}^{-2}}]$

$V_O=27\times {10}^{5}V$