Question

How many electrons must be removed from each of two 4.85-kg copper spheres to make the electric force of repulsion between them equal in magnitude to the gravitational attraction between them?

Answer 1

`n=2.611*10^(9)electrons`

Step-by-step explanation:

Given data

mass of copper m=4.85 kg

charge of electron qe= -1.6×10⁻¹⁹C

To find

Number of electron n must be removed

Solution

Equate the magnitude of electric force Fe of repulsion between two two spheres to the magnitude of gravitational force of attraction between them

So

`F_(e)=F_(g)\\ k(q_(e)q_(e))/(r^(2) )=G(m^(2) )/(r^(2) )\\ k(q_(e))^(2)=Gm^(2)\\`

where q is charge of each sphere which is equal to number n of removed electrons multiplied by each charge qe

So

`k(nq_(e))^(2)=Gm^(2)\\n=\sqrt{(Gm^(2))/(k(q_(e))^(2)) }\\ n=\sqrt{((6.67*10^(-11)N.m^(2)/kg^(2))(4.85kg)^(2))/((8.99*10^(9)N.m^(2)/C^(2) )(1.6*10^(-16)C)^(2)) }\\n=2.611*10^(9)electrons`