Question

A regular hexagon of side $10cm$ has charge $5\mu c$ each at each of its vertices. Calculate the potential at the centre of the hexagon.

Answer 1

ABCDEF is a regular hexagon of side 10 cm each. At each corner, the charge $q=5\mu C$ is placed. O is the center of the hexagon

Given $AB=BC=CD=DE=EF=FA=10cm$. As the hexagon has six equilateral triangles so the distance of center O from every vertex is 10 cm.

i.e. $OA=OB=OC=OD=OE=OF=10cm$.

Potential at point O = Sum of potential at center O due to individual point charge

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

As $V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$

then

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

Putting the values, we get

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

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

$V_O=2.7\times {10}^{6}V$