Elecrticity and coulomb law?

Question

2 balls with a mass of 100kg are at a distance of 5m from one another, how many electrons need to be moved from one to the other in order to balance the gravitational force by an electric force,

from what i understand, we have 2 neutral balls of the same mass, and i need to find out how much electric charge is needed to balance one of these balls with mg, i suppose the other is attached to the ceiling or something to prevent it from falling.

whatever the ones charge is the others will be its negative

mg=[k(q1*q2)]/r^2

since q1 and q2 are the same

mg=[k(q^2)]/r^2

q^2=(mg)(r^2/k)

q=[(mg)(r^2/k)]^0.5

q^2=[(100*9.8)(25/9*10^9)]=2.72*10^-6

q=1.65*10^-3 c

now q is the charge of the balls, positive and negative,

what i must do now is divide my total charge by the charge of a single electron (1.6*10^-19) and ill get the amount of electrons, only this doesnt seem to be working for me, where have i gone wrong??

the final answer is 5.38*10^10 electrons

Answer 1

I understand the question a bit differently. I think the problem  stresses the attraction due to gravity between the balls not the attraction between the Earth and the balls.

So

Fe= Fg

k(q1*q2)]/R^2 = G m1 m2 / R^2

k= electrostatic constant

G - gravitational constant

Finally since m1 = m2 and q1=q2 we have teh charge required to be placed on each ball

q = sqrt [(G/k) m^2]  and

Number of electrons to be place on each ball is

N= q/1.602 x 10^-19

Answer 2

try it as follows

First calculate the gravitational force by the standard newton's formula.

Gm1m2/r^2.

you are taking it as mg, but g=GM/r^2, and it is 9.8 only for mass of earth. here you have to consider mass of balls and distance between them.

then equate it with coulombs force.

i have a doubt though, gravitational force is always attractive, so to balance it we need repulsive coulomb force, but in this case it will come as attractive.(opp charges will be formed)

nevertheless try to calculate it, magnitude will actually be the same.