Question

In Millikan’s oil-drop experiment, one looks at a small oil drop held motionless between two plates. Take the voltage between the plates to be 2210 V, and the plate separation to be 1.87 cm. The oil drop (of density 0.816 g/cm^3) has a diameter of 3.3x 10^−6 m . Calculate the charge in terms of the number of elementary charges (1.6 x 10^-19).

Answer 1

Answer:

q = 1,297 10⁻¹⁹ C , n=1

Explanation:

For this problem we will use Newton's second law in the case of equilibrium

        $F_{e}$ + B - W = 0

Where $F_{e}$ is the electrical force up, B the thrust and W the weight of the drop.

Let's look for weight and thrust

oil

      ρ = m / V

      m =  ρ V  

Air

       B =   $\rho _{air}$ g V

Electric force

       $F_{e}$ = qE

       E = V / d

       $F_{e}$ = q V/d

Let's replace

      q V / d + $\rho _{air}$ g V - ρ V g = 0

      qV / d = (4/3 π r³) g (ρ –$\rho _{air}$)

      q = 4/3 π r³ (ρ –$\rho _{air}$)  d / V

Reduce to SI units

      d = 1.87 cm (1m / 100cm) = 1.87 10⁻² m

      ρ= 0.816 r / cm3 (1kg / 1000g) (102cm / 1m)³ = 816 kg / m³

      $\rho_{air}$ = 1.28 kg / m³

Let's calculate the charge

       r = d / 2 = 3.3 10⁻⁶ m

      r = 1.65 10⁻⁶ m

     q = 4/3 π (1.65 10⁻⁶)³  (816 - 1.28)   0.0187 / 2210

     q = 12.9717 10⁻²⁰ C

     q = 1,297 10⁻¹⁹ C

If we assume that the load is

     q = n e

In this case n = 1