Answer:
891 excess electrons must be present on each sphere
Explanation:
One Charge = q1 = q
Force = F = 4.57*10^-21 N
Other charge = q2 =q
Distance = r = 20 cm = 0.2 m
permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)
Using Coulomb's law,
F=[1/4pieo]q1q2/r^2
F = [1/4pieo]q^2 / r^2
q^2 =F [4pieo]r^2
q = r*sq rt F[4pieo]
q=0.2* sq rt[ 4.57 x 10^-21]*[4*3.1416*8.854*10^-12]
q = 1.42614*10^ -16 C
number of electrons = n = q/e=1.42614*10^ -16 /1.6*10^-19
n =891
891 excess electrons must be present on each sphere
