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.
1 answer:
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, = -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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