Question

In a Millikan experiment, a droplet of mass 4.7 x 10^-15 kg floats in an electric field of 3.20 x 104 N/C.

Answer 1

Answer:

a. The force of gravity on the droplet is approximately 4.606 × 10⁻¹⁴ Newtons

b. The electric force that balances the force of gravity is approximately -4.606 × 10⁻¹⁴ Newtons

c. The excess charge is approximately -1.439375 × 10⁻¹⁸ C

d. There are approximately 9 excess electrons on the droplet

Explanation:

The parameters of the Millikan experiment are;

The mass of the droplet, m = 4.7 × 10⁻¹⁵ kg

The electric field in which the droplet floats, E = 3.20 × 10⁴ N/C

a. The force of gravity on the droplet, F = The weight of the droplet, W = m × g

Where;

g = The acceleration due to gravity ≈ 9.8 m/s²

W = 4.7 × 10⁻¹⁵ kg × 9.8 m/s² = 4.606 × 10⁻¹⁴ Newtons

∴ The force of gravity on the droplet = W = 4.606 × 10⁻¹⁴ Newtons

b. The electric force that balances the force of gravity, $F_v$ = -W = -4.606 × 10⁻¹⁴ Newtons

c. The excess charge is given as follows;

$F_v$ = q·E

∴ The electric force that balances the force of gravity, $F_v$ = q·E =  -4.606 × 10⁻¹⁴ N

q·E =  -4.606 × 10⁻¹⁴ N

q =  -4.606 × 10⁻¹⁴ N/E

∴ q = -4.606 × 10⁻¹⁴ N/(3.20 × 10⁴ N/C) ≈ -1.439375 × 10⁻¹⁸ C

The excess charge, q ≈ -1.439375 × 10⁻¹⁸ C

d. The charge of one electron, e = 1.602176634 × 10⁻¹⁹C

The number of excess electrons in the droplet, n, is given as follows;

n = 1.439375 × 10⁻¹⁸ C/(1.602176634 × 10⁻¹⁹C) = 8.98387212405 electrons

∴ n ≈ 9 electrons.