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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?`
8 years ago

Pratham Ashish
17 Points
```
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

```
8 years ago
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