A charge Q is spread uniformly in the form of a line charge density λ=Q/3a on the sides of an equilateral triangle 3a. Calculate the potential at the centroid C of the triangle?

let us take the arm AB,,

  dV = 1/ 4¶ε  dq / r

  if M is the mid point of AB . then MG = (√3/ 2)  ( 3a )  1/3  = √3 a / 2    & if l is the distance of dQ from M ,

 l =  √3 a / 2  / ( tan x )

 dl = - √3 a / 2 cosec ^2 x dx

 r =( √3 a / 2  ) / sin x

 dV = ( 1/ 4¶ε ) dq . sin x  ( 2/√3 a  )

     dq =  λ dl = - λ  (√3 a / 2) cosec ^2x dx

 dV =  ( 1/ 4¶ε )  ( 2/√3 a  ) sin x .  λ  (√3 a / 2) cosec ^2x dx

dV =  ( 1/ 4¶ε )  λ   cosec x dx

  V  = 2 ∫  ( 1/ 4¶ε )  λ   cosec x dx     from x= 30 to x= 90

  V =  ( 1/ 2¶ε )  λ  ln[ cosecx - cot x ]           from x= 30 to x= 90

     =  ( 1/ 4¶ε )  λ   ln [ 2- √3 ]

total potential =  (3 / 4¶ε )  λ   ln [ 2- √3 ]

                     =  Q  ln [ 2- √3 ] / 4¶εa