# Two small spheres spaced 20.0 centimeters apart have equal charge

Question 1

Two small spheres spaced 20.0 centimeters apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57*10−21−21 newtons?

Question 2

Two small spheres spaced 20.0 centimeters apart have equal charge.How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is newtons?

Question 3

What careers are associated with Newton’s law of motion?

Given:

• Separation between the spheres: d = 20 cm = 0.2 md = 20 cm = 0.2 m
• Magnitude of the force between the spheres: F = 4.57×10−21F = 4.57×10−21

Let Q = −qQ = −q be the charge on each sphere. From Coulomb’s law, force of repulsion between the spheres is:

F = k×q2d2F = k×q2d2

Substituting the values,

4.57×10−21 = 9×109×q20.22⟹ q = 1.425×10−16 C

The force between two charge sphere can be approximately calculateby the equation.
E=kq^2/r^2
so q=sqrt(Er^2/k)=1,425e-16(C)
the charge of one electron is 1,6e-19 (C)
so the number of electron is 1,425e-16/1,6e-19=891

n =891

891 excess electrons must be present on each sphere