The correct option (C) (a/√2)
Explanation:
![](https://www.sarthaks.com/?qa=blob&qa_blobid=16309278133955325079)
F1 = [(kq0q)/(a2 + y2)] & F2 = [(kq0q)/(a2 + y2)]
∴ F1 = F2
by symmetry x component of force will cancel each while y axis, the
components will add up.
∴ The resultant force + q0 is
F = F1 cos θ + F2 cos θ
= 2F1 cos θ F1 = F2
F = [(2kq0q)/(a2 + y2)]cos θ (1)
and cos θ = [y/√(a2 + y2)]
from (1) F = [(2kq0qy) / {(a2 + y2)(3/2)}] = 2kqq0y(a2 + y2)–(3/2)
Force on charge q0 will maximum when (dF/dy) = 0
∴ 2kq0q [{1/(a2 + y2)(3/2)} + y [– (3/2)] (a2 + y2)–(5/2)(2y)] = 0
= 2kqq0 [(a2 + y2)–(3/2) – 3y2(a2 + y2)–(5/2)] = 0
∴ (a2 + y2)–(3/2) = 3y2(a2 + y2)–(5/2)
∴ a2 + y2 = 3y2
∴ 2y2 = a2
∴ y = (a/√2)