Question

How much equal charge should be placed on the earth and the moon so that the electrical repulsion balances the gravitational force of 1.98×1020n? Treat the earth and moon as point charges a distance 3.84×108m apart.

Answer 1

As we know that electrostatic force between two charges is given as

`F = (kq_1q_2)/(r^2)`

here we know that electrostatic repulsion force is balanced by the gravitational force between them

so here force of attraction due to gravitation is given as

`F_g = 1.98 * 10^(20) N`

here we can assume that both will have equal charge of magnitude "q"

now we have

`1.98 * 10^(20) = (kq^2)/(r^2)`

`1.98 * 10^(20) = ((9* 10^9)(q^2))/((3.84 * 10^8)^2)`

`1.98 * 10^(20) = (6.10 * 10^(-8)) q^2`

now we have

`q = 5.7 * 10^(13) C`