Question

A small sphere whose mass is 0.1gm carries a charge of 3 × 10^–10C and is tie up to one end of a silk fiber 5cm long. The other end of the fiber is attached to a large vertical conducting plate which has a surface charge of 25 × 10^–6Cm^–2, on each side. When system is freely hanging the angle fiber makes with vertical is ……… 

(A) 41.8° 

(B) 45° 

(C) 40.2° 

(D) 45.8°

Answer 1

The correct option (C) 40.2°

Explanation:

m = 0.1 × 10^–3 kg

q = 3 × 10^–10C

ρ = surface charge density = 25 × 10^–6 C/m²

from figure :  T sin θ = qE

T cos θ = mg

∴ Taking ratio, we get tan θ = (qE/mg)   (1)

Also E = (ρ/∈₀) = [(25 × 10^–6)/(8.85 × 10^–12)] = 2.82 × 10⁶

∴ from (1),

 tan θ  = [(3 × 10^–10 × 2.82 × 10⁶)/(0.1 × 10^–3 × 10)]

= 8.47 × 10^–4 × 10³_­

tan θ  = 0.847

θ = tan^–1 (0.847) = 40.2°