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, [tex]F_v[/tex] = -W = -4.606 × 10⁻¹⁴ Newtons
c. The excess charge is given as follows;
[tex]F_v[/tex] = q·E
∴ The electric force that balances the force of gravity, [tex]F_v[/tex] = 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.
a. The force of gravity on this droplet is [tex]4.606 \times 10^{-14} \;Newton[/tex]
b. The electric force that balances this droplet is [tex]-4.606 \times 10^{-14} \;Newton[/tex]
c. The excess charge is equal to [tex]-1.44 \times 10^{-18}\;C[/tex]
d. The number of excess electrons that are on this droplet is 9.0 electrons.
Given the following data:
Scientific data:
a. To calculate the force of gravity on this droplet:
The force of gravity on this droplet is equal to the weight of the droplet and it would be calculated by using this formula:
[tex]F = mg\\ \\ F = 4.7 \times 10^{-15} \times 9.8\\ \\ F = 4.606 \times 10^{-14} \;Newton[/tex]
b. To calculate the electric force that balances this droplet:
The electric force that balances this droplet is equal to the negative weight of the droplet.
[tex]F_e = -mg = -4.606 \times 10^{-14} \;Newton[/tex]
c. To calculate the excess charge:
Mathematically, the excess charge is given by this formula:
[tex]q=\frac{F_e}{E} \\ \\ q=\frac{-4.606 \times 10^{-14}}{3.2 \times 10^4} \\ \\ q=-1.44 \times 10^{-18}\;C[/tex]
d. To calculate the number of excess electrons that are on this droplet:
[tex]n=\frac{1.44 \times 10^{-18}}{1.6 \times 10^{-19}} [/tex]
n = 9.0 electrons
Read more on electric field here: https://brainly.com/question/14372859
