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. a. What is the force of gravity on this droplet?

b. What is the electric force that balances it?

c. What is the excess charge?

d. How many excess electrons are there on this droplet?

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, = -W = -4.606 × 10⁻¹⁴ Newtons

c. The excess charge is given as follows;

= q·E

∴ The electric force that balances the force of gravity, = 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.

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