Answer:
[tex]n=2.611*10^{9}electrons[/tex]
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
[tex]F_{e}=F_{g}\\ k\frac{q_{e}q_{e}}{r^{2} }=G\frac{m^{2} }{r^{2} }\\ k(q_{e})^{2}=Gm^{2}\\[/tex]
where q is charge of each sphere which is equal to number n of removed electrons multiplied by each charge qe
So
[tex]k(nq_{e})^{2}=Gm^{2}\\n=\sqrt{\frac{Gm^{2}}{k(q_{e})^{2}} }\\ n=\sqrt{\frac{(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[/tex]
