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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?

Sagot :

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.

Lanuel

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:

  • Mass of droplet = [tex]4.7 \times 10^-15 \;kg[/tex]
  • Electric field = [tex]3.20 \times 10^4 \;N/C.[/tex]

Scientific data:

  • Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]
  • Charge of electron = [tex]1.6 \times 10^{-19}\;C[/tex]

The parameters of an electric field.

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

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