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?

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.