A regular hexagon of side 10 cm. has a charge of `5muC` at each of its corners. Calculate the potential at the centre of hexagon.

Explanation

ABCDEF is a regular hexagon of side 10 cm each. At each corner, the charge `q=5muC` is placed. O is the centre of the hexagon.

Given, `AB=BC=CD=DE=EF=FA=10` cm
As. The hexago has six equilateral triangles, so the distance of centre O from every vertex is 10 cm.
i.e. `OA=OB=OC=OD=OE=OF=10 cm`
Potential at point O = sum of potential at centre O due to individual point charge
`:.V_(O)=V_(A)+V_(B)+V_(C)+V_(D)+V_(E)+V_(F)`
`V_(O)=(1)/(4piepsi_(0))*[(q)/(OA)+(q)/(OB)+(q)/(OC)+(q)/(OD)+(q)/(OE)+(q)/(OF)]`
`(becauseV=-(1)/(4pi epsi_(0))*q/r)`
Putting the values, we get
`V_(O)=9xx10^(9)[(5xx10^(-6))/(10xx10^(-2))+(5xx10^(-6))/(10xx10^(-2))+(5xx10^(-6))/(10xx10^(-2))+(5xx10^(-6))/(10xx10^(-2))+(5xx10^(-6))/(10xx10^(-2))+(5xx10^(-6))/(10xx10^(-2))]`
`=9xx10^(9)xx(6xx10^(-6)xx5)/(10xx10^(-2))=27xx10^(5)`
`:.V_(O)=2.7xx10^(6)V`