#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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?

12 years ago 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