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An electron is in a vacuum near the surface of the Earth. Where should a second electron be placed so that the net force on the first electron to the other electron and to gravity, is zero?

2 Answers

6 votes

Final answer:

To achieve a zero net force on an electron near Earth's surface, a second electron needs to be placed above it. The electrostatic repulsion balances the electron's weight due to gravity, highlighting the considerable difference in strength between gravitational and electrostatic forces.

Step-by-step explanation:

The question is seeking to understand the balance between electrostatic and gravitational forces acting on an electron near the surface of the Earth. We know that the electrostatic force between two electrons is repulsive, and gravity pulls objects towards Earth's center. To balance these forces so that the net force on the first electron is zero, the second electron must be placed above the first electron. This is because gravity will pull the first electron downwards and the electrostatic repulsion will push it upwards. To achieve a net force of zero, the repulsion must counteract the weight of the first electron.

The small value of the electric field needed to support an electron against Earth's gravity demonstrates the vast difference in strength of these forces — the gravitational force is comparatively negligible against the electrostatic force. This disparity can be seen in applications such as cathode-ray tubes, where relatively small electric fields can produce significant accelerations in electrons, as opposed to the effects of gravity.

User Dboswell
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There are two forces acting on the first electron.

1) Force of gravity, Fg = mass of the electron * g

mass of the electron = 9.11 * 10 ^ -31 kg
g = 9.8 m/s^2

Fg = 9.11 * 10^ -31 kg * 9.8 m/s^2

2) Electrostatic force due to the second electron, Fe

Use Coulomb Law.

Fe = Coulomb constant * [charge of the electron*charge of the electron] / [separation]^2

Coulomb constant = 9.0*10^9 N*m^2 /C^2
Charge of the electron = 1.6 * 10 ^-19 C
Separation = d

Fe = 9.0 * 10^9 N*m^2/C^2 * [1.6*10^-19C]^2 / d^2

Condition: net force = 0 ==> Fe = Fg

Given that the second electron will exert a repulsion force, it has to be below (closer to the earth than) the first electron to counteract the atractive force of the earth.

9.11 * 10^ -31 kg * 9.8 m/s^2 = 9.0 * 10^9 N*m^2/C^2 * [1.6*10^-19C]^2 / d^2

From which you can solve for d.

d = sqrt { 9.0 * 10^9 N*m^2/C^2 * [1.6*10^-19C]^2 / (9.11 * 10^ -31 kg * 9.8 m/s^2)}

d = 5.08 m

Then the second electron must be placed 5.08 m below the first electron.







User Dmitriy Khaykin
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