Three equal charges Q are placed at the three corners of an equilateral triangle of side length a. Work done in shifting a charge q from infinity to the centroid of the triangle is

Options

A. `[1/(6piepsilon_0)q^2/(k)]^(1//2)`
B. `[sqrt3/(12piepsilon_0)q^2/(k)]^(1//2)`
C. `[sqrt3/(12piepsilon_0)q^2/(k)]^(1//2)`
D. `[sqrt3/(4piepsilon_0)q^2/(k)]^(1//2)`

Answer

B

Explanation

F(r) = kr
Now `F_("net")` on a particle is `2F_q cos 30^@` due to the other two charges
`F_("net") = (2kq^2)/(a^2) xxsqrt3/2 `
Also, `r=2/3 (sqrt3/2a)`
`:.a=sqrt3r` replacing it in `F_("net")` we get
`F_("net") = (2kq^2)/((sqrt3r)^2)xx(sqrt3/2)=(kq^2)/(sqrt3r^2)`
This is balanced by F(r)
`:. F(r) =n F_("net") implies kr =(1xxq^2)/(4piepsilon_0xxsqrt3r^2)`
`:. r = ((sqrt3q^2)/(12 pi epsilon_0k))^(1//3)`