Question

Two point positive charges q each are placed at (– a, 0) and (a, 0). A third positive charge q₀ is placed at (0, y). For which value of y the force at q₀ is maximum ……… 

(A) a 

(B) 2a 

(C) (a/√2) 

(D) (a/√3)

Answer 1

The correct option (C) (a/√2)   

Explanation:

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)