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?
Answer to question 1
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
Answer to question 2
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
Answer to question 3
Some careers that are associated with Newton’s laws of motion are engineers, astronauts, and physicists.
In each of these careers, the three laws of motions are fundamental in ensuring that the quality of work, the amount of expenditure and the safety of the workers will be optimized.