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 electroncharge of the electron] / [separation]^2
Coulomb constant = 9.010^9 Nm^2 /C^2
Charge of the electron = 1.6 10 ^-19 C
Separation = d
Fe = 9.0 10^9 Nm^2/C^2 [1.610^-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 Nm^2/C^2 [1.610^-19C]^2 / d^2
From which you can solve for d.
d = sqrt { 9.0 10^9 Nm^2/C^2 [1.610^-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.
