a charge q is spread uniformly in form of line charge density q/3a on sides of equilateral triangle of side 3a. calculate potential at centroid .an:2.634qk/a

Dear shefali

potential is a scalor quantity so potential due to whole triangle will be equal to 6 timen the potential due to portion BD

let a point P at a distance x from B

 dq = q/3a dx

PO =√(PD² +DO²)

     =√(3a/2 -x)² + (3a/2√3)²)

potenial due to charge dq at the center is

  dV= kdq/PO

  dV   =k q/3a dx/{√(3a/2 -x)² + (3a/2√3)²}

  V = _o∫^3a/2 kq dx /3a{√(3a/2 -x)² + (3a/2√3)²}

     =kq/3a  _o∫^3a/2  dx /{√(3a/2 -x)² + (3a/2√3)²}

V = kq/3a [ log |x-3a/2  + {√(3a/2 -x)² + (3a/2√3)²}|]  limit o to 3a/2

  =kq/3a [ log |3a/2 -3a/2  + {√(3a/2 -3a/2 )² + (3a/2√3)²}|- log |0-3a/2  + {√(3a/2 -0)² + (3a/2√3)²}|]

  = kq/3a [ log |(3a/2√3)|- log |-3a/2  + {√(3a/2 )² + (3a/2√3)²}|]

  =-kq/3a log(2-√3)

total potential =6V =-2kq/a log(2-√3)

                           =2.63kq/a

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