Answer to Question #153504 in Electricity and Magnetism for Lila

Question

Suppose earth and moon carry + charges of equal magnetude. How large would charge need to be to produce an electrostatic repulsion equal 1% of gravitational attraction between the 2 bodies?

Answer 1

Equate the one percent of gravitational force between the moon and Earth to Coulomb force between the moon and the Earth and find the charge on each body.

Gravitational force for Earth and moon system:

$F_G = \frac{GMm}{R^2}$

1 % for $F_G$ is

$F_{1\%}=\frac{F_G \times 1}{100} \\ = \frac{\frac{GMm}{R^2} \times 1}{100} \\ F_{1\%}=\frac{GMm}{100R^2}$

M = mass of Earth

m = mass of Moon

R = distance between them

Let Q is the charge on moon or earth, separated by the distance R.

Coulomb’s force for any two charges separated by a distance R is

$F_C = k \frac{QQ}{R^2} \\ F_{1\%}=F_C \\ \frac{GMm}{100R^2} = k \frac{QQ}{R^2} \\ \frac{GMm}{100} = kQ^2 \\ Q = \sqrt{\frac{GMm}{100k}} \\ = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.36 \times 10^{22}}{100 \times 8.99 \times 10^9}} \\ = \sqrt{32.6 \times 10^{24}} \\ = 5.71 \times 10^{12} \;C \\ ≈ 6 \times 10^{12} \;C$