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

Answer:

$n=2.611*10^{9}electrons$

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

$F_{e}=F_{g}\\ k\frac{q_{e}q_{e}}{r^{2} }=G\frac{m^{2} }{r^{2} }\\ k(q_{e})^{2}=Gm^{2}$

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

So

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