Equal charges are deposited on two spherical bodies, each having a...

The Coulomb repulsion force between the two charges is given by $F_e = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{r^2}$

The gravitational attraction force between the two masses is given by $F_g = G \dfrac{m^2}{r^2}$

Since the Coulomb repulsion exactly balances the gravitational attraction, we have $F_e = F_g$

$\dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{r^2} = G \dfrac{m^2}{r^2}$

$q^2 = 4\pi\epsilon_0 G m^2$

$q = m \sqrt{4\pi\epsilon_0 G}$

Substituting the given values $m = 50 \text{ kg}$, $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$, and $\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$:

$q = 50 \times \sqrt{\dfrac{6.67 \times 10^{-11}}{9 \times 10^9}}$

$q = 50 \times \sqrt{0.741 \times 10^{-20}}$

$q \approx 50 \times 0.86 \times 10^{-10}$

$q \approx 43 \times 10^{-10} \text{ C} = 4.3 \times 10^{-9} \text{ C}$