Question

Millikan used tiny oil droplets and adjusted the electric field until the electric force balanced the weight of the droplet, and one of the droplets hovered in air. However, the drops are all different sizes and weights, and the amount of charge on each droplet also varies. So, the tuning of the electric field is different for droplets of different sizes, which carry different charges. Let us assume that the density of the oil Millikan used in his experiment is 800 kg m3

A. Calculate the mass (in kg) of the hovering droplet if its radius is measured to be 65 um.
B. Calculate the electric field (in V/m) required to counter the weight of the above droplet if it carries a charge q = 2e.
C. In the second trial, the hovering droplet has a mass of 10-15 kg. It carries a charge of q = 11e. calculate the electric field (in V/m) required to counter the weight of this droplet.

Answer 1

A. 9.203 × 10⁻¹⁰ kg

B. 2.82 × 10¹⁰ V/m

C. 5.56 × 10³ V/m

Step-by-step explanation:

A. Calculate the mass (in kg) of the hovering droplet if its radius is measured to be 65 um.

We know density, ρ = mass of oil drop, m/volume of oil drop, v

m = ρv

ρ = 800 kg/m³ and v = 4πr³/3 (ince the oil drop is a sphere) where r = radius of oil drop = 65 μm = 65 × 10⁻⁶ m.

So, m = ρv = ρ4πr³/3

= 800 kg/m³ × 4π(65 × 10⁻⁶ m)³/3

= 800 kg/m³ × 4π274625 × 10⁻¹⁸ m³/3

= 878800000π × 10⁻¹⁸ kg/3

= 2760831623.97 × 10⁻¹⁸ kg/3

= 920277207.99 × 10⁻¹⁸ kg

= 9.203 × 10⁻¹⁰ kg

B. Calculate the electric field (in V/m) required to counter the weight of the above droplet if it carries a charge q = 2e.

Since the electric force F = qE where q = charge on oil drop = 2e where e = electron charge = 1.602 × 10⁻¹⁹ C and E = electric field equals the weight of the oil drop W = mg where m = mass of oil drop = 9.203 × 10⁻¹⁰ kg and g = acceleration due to gravity = 9.8 m/s².

So, F = W

qE = mg

E = mg/q

E = mg/2e

substituting the values of the variables into the equation, we have

E = 9.203 × 10⁻¹⁰ kg × 9.8 m/s²/(2 × 1.602 × 10⁻¹⁹ C)

E = 90.1894 × 10⁻¹⁰ kg-m/s²/3.204 × 10⁻¹⁹ C

E = 28.15 × 10⁹ V/m

E = 2.815 × 10¹⁰ V/m

E ≅ 2.82 × 10¹⁰ V/m

C. In the second trial, the hovering droplet has a mass of 10-15 kg. It carries a charge of q = 11e. calculate the electric field (in V/m) required to counter the weight of this droplet.

Since F = W

qE = mg

E = mg/q

E = mg/11e

E = 10⁻¹⁵ kg × 9.8 m/s²/(11 × 1.602 × 10⁻¹⁹ C)

E = 9.8 × 10⁻¹⁵ kg-m/s²/17.622 × 10⁻¹⁹ C

E = 0.556 × 10⁴ V/m

E = 5.56 × 10³ V/m