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Ainat [17]
2 years ago
9

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?
Physics
1 answer:
Fiesta28 [93]2 years ago
6 0
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.


 




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