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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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