Question

A positively charged oil drop of mass is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced 16 cm apart. If the mass of the drop is 8.0 10-15 kg and it remains stationary when the potential difference between the plates is 2.44 kV, what is the magnitude of the charge on the drop

Answer 1

the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸ C

Step-by-step explanation:

Given that;

mass of the drop m = 8.0×10⁻¹⁵ kg

distance d = 16 cm = 0.16 m

potential difference between the plates V = 2.44 kV = 2440 v

acceleration of gravity g = 9.81 m/s²

the magnitude of the charge on the drop = ?

weight is balanced by the electrostatic force

weight = mg = 8.0×10⁻¹⁵ kg × 9.81 m/s² = 7.848 × 10⁻¹⁴

we know that; V = Ed

E = V/d = 2440 / 0.16 = 15200 v/m

Electrostatic force = qE

so weight = qE

q = weight / E

q = 7.848 × 10⁻¹⁴ / 15200

q = 5.163 × 10⁻¹⁸ C

Therefore, the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸ C