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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
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
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