Answer:
The charge on the oil drop is 3e
Explanation:
F = qE
Where;
F is the applied force in Newton
E is the electric field potential N/C
q is charge in C
Given;
Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m
density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³
Electric field potential = 1.92 ✕ 10⁵ N/C
F =mg
mass of the oil drop = density, ρ X volume of the oil drop
volume of the oil drop (spherical) = (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³
⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³
mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)
⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg
⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N
F = qE
q = F/E
q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C
In terms of e
1e = 1.6 X10⁻¹⁹ C
= (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e
Therefore, the charge on the oil drop is 3e
