A charged dust particle of radius 5×10⁻⁷ m is in a horizontal electric field of intensity 6.28×10⁵ V m⁻¹. Air viscosity is η = 1.6×10⁻⁵ N s m⁻². If its uniform horizontal speed is 0.02 m s⁻¹, the number of electrons on it is k×10. Find k.
Physics · Mixed Chapter
JEE Mains

Question

A charged dust particle of radius 5×10⁻⁷ m is in a horizontal electric field of intensity 6.28×10⁵ V m⁻¹. Air viscosity is η = 1.6×10⁻⁵ N s m⁻². If its uniform horizontal speed is 0.02 m s⁻¹, the number of electrons on it is k×10. Find k.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Forces at uniform speed: At terminal horizontal velocity, acceleration is zero. The electric force balances viscous drag: |q|E = 6πηrv, with |q| = Nₑe. Step 2 — Calculate number of electrons: Nₑ = 6πηrv/(eE). Insert η = 1.6×10⁻⁵ N s m⁻², r = 5×10⁻⁷ m, v = 0.02 m s⁻¹, E = 6.28×10⁵ N C⁻¹, e = 1.6×10⁻¹⁹ C. Nₑ ≈ 3.0×10¹. Step 3 — Identify requested integer: If Nₑ = k×10, then k = 3. Answer: 3.