Question

A shower of protons from outer space deposits equal charges +q on the earth and the moon when electrostatic repulsion exactly counterbalances the gravitational attraction. How large is q?

Answer 1

Step 1: Given data

1. The net charge of the proton shower is $+q$.
2. The gravitational attraction and electrostatic repulsion of earth and moon are equal in this case.

Step 2: Formula used

1. The gravitational attraction between the earth and moon is defined by the form, $F_G=\frac{GMm}{r^2}$, where, r is the distance between earth and moon, G is the universal gravitational constant, and M and m are the mass of the earth, and moon.
2. The electrostatics repulsive force between two charges $q_1$ and $q_2$ is defined by the form, $F_e=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{q_1\times q_2}{r^2}$, where, r is the distance between the charges, $q_1$ and $q_2$, $k=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}$is the Coulomb constant.

Step 3: Finding the charge q

According to the question, the electrostatic repulsion of earth and moon is the same as the gravitational attraction between earth and moon.

So,

$$\begin{gathered} \frac{GMm}{r^2}=k\frac{q\times q}{r^2}(\sin ce,q_1=q_2=q) \\ or{q}^2=\frac{GMm}{k} \\ orq=\sqrt{\left(\frac{GMm}{k}\right)} \end{gathered}$$

Therefore, the amount of q is $\sqrt{\left(\frac{GMm}{k}\right)}$.