How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−21 newtons?

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  

it is the second option.

.) the observation by semmelweis that more mothers died of childbed fever when their doctors also worked in the morgue