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Question:

A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength 8.1π × 10⁵ Vm⁻¹. When the field is switched off, the drop is observed to fall with terminal velocity 2 × 10⁻⁷ ms⁻¹. Given g = 9.8 ms⁻², viscosity of the air = 1.8 × 10⁻⁵ Nsm⁻², and the density of oil = 900 kgm⁻³, the magnitude of q is?

8.0 × 10⁻⁹ C

1.6 × 10⁻⁹ C

3.2 × 10⁻⁹ C

4.8 × 10⁻⁹ C

Solution:

Here, E = 8.1π × 10⁵ Vm⁻¹, v = 2 × 10⁻⁷ ms⁻¹, η = 1.8 × 10⁻⁵ Nsm⁻²; ρ = 900 kgm⁻³

When the electric field is switched off, let the drop falls with terminal velocity v, then
v = (2r²/9η)(ρ - σ)g
or
r = [9ηv/2(ρ - σ)g]¹/²
∴ q = 1E × (4/3)πρg[9ηv/2(ρ - σ)g]³/²
= (4/3) × 8.1π × 10⁵ × π × 900 × 9.8 × [9 × 1.8 × 10⁻⁵ × 2 × 10⁻⁷/2 × 900 × 9.8]³/²
On solving we get,
q = 8 × 10⁻⁹ C