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