Two small spheres spaced 20.0 cm apart have equal charge. part a how many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33×10−21 n?

As the each sphere have the same charge, we use the Coulomb law to calculate the charge on the sphere

Here, and and k is constant and its value is  

Therefore,

Thus, the number of electron present in each sphere is

Solution:

Since we have two small spheres:

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[3.33 x 10^-21]*[4*3.1416*8.854*10^-12]  

q = 1.28923*10^ -16 C  

So,

number of excess electrons = n = q/e=1.28923*10^ -16 /1.6*10^-19  

n =750  

Thus,

750 excess electrons must be present on each sphere.

grav potential energy

there are no contact forces mentioned on the list of choices that you've provided.