 # 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 Badiuddin askIITians.ismu Expert
148 Points
12 years ago

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 = o3a/2 kq dx /3a{√(3a/2 -x)2 + (3a/2√3)2}

=kq/3a  o3a/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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