Question

Two identical charged spheres suspended from a common point by two mass less strings of length ℓ are initially a distance d(d << ℓ) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the spheres approach each other with a velocity ѵ. Then function of distance x between them becomes ……..

(A) ѵ ∝ x 

(B) ѵ ∝ x^(–1/2) 

(C) ѵ ∝ x^–1 

(D) ѵ ∝ x^(1/2)

Answer 1

The correct option (B) ѵ ∝ x^(–1/2)   

Explanation:

For equilibrium,

T cos θ = mg     (1)

T sin θ = Fe = [(k ∙ q ∙ q) / x²]      coulomb's law

 i.e. T sin θ = [(kq²)/x²]   (2)

 Dividing (2) by (1)

[(T sin θ)/(T cos θ)] = [(kq²) / (x² ∙ mg)]

∴ tan θ = [(kq²) / (x²mg)] also tan θ = [(x/2)/ℓ] = (x/2ℓ)

∴ (x/2ℓ) = [(kq²)/(x²mg)]

∴ q² = [(x³mg)/(2kℓ)]   (2)

differential (2) wrt t.

∴ 2q(dq/dt) = [(mg)/(2kℓ)] × 3x² ∙ (dx/dt)

∴ 2q(dq/dt) = 3x²(dx/dt) × (q²/x³)      from (2)

∴ 2(dq/dt) = (q/x)3 ∙ (dx/dt)

∴ (2/3) (dq/dt) = (ѵ/x)[x^(3/2){(mg)/(2kℓ)^(1/2)}]

∴ [{(2/3) (dq/dt)}/{(mg)/(2kℓ)^(1/2)}] = ѵx^(1/2)

∴ ѵx^(1/2) = constant

hence ѵ ∝ x^(–1/2)