Suppose you had two small boxes, each containing 1.0 g of protons. (a) If one were placed on the moon by an astronaut and the other were left on the earth, and if they were connected by a very light (and very long!) string, what would be the tension in the string? Express your answer in newtons and in pounds. Do you need to take into account the gravitational forces of the earth and moon on the protons? Why? (b) What gravitational force would each box of protons exert on the other box?

According to the Coulomb’s law, the force between two charges spaced at a distance is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.

Coulomb’s law of force between two charges is,

$F=k\frac{\left|q_1{q}_2\right|}{r^2}$ ..... (1)

Here, the Coulomb’s constant, $k=8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2,r$is the distance between two charges $q_1$and $q_2$.

(a) Determination of the tension in the string.

The data required in this problem are,

Mass of box, $m=1.00\times {10}^{-3}\mathrm{kg}$

Mass of proton, $m_p=1.67\times {10}^{-27}\mathrm{kg}$

Charge of a proton, $e=1.60\times {10}^{-19}\mathrm{C}$.

The distance between the earth and the moon, $r=3.84\times {10}^8\mathrm{m}$.

To determine the total charge, the number of atoms in the box is required. Therefore

$$\begin{gathered} N=\frac{m}{m_p} \\ =\frac{1.0\times {10}^{-3}kg}{1.67\times {10}^{-27}kg} \\ =5.99\times {10}^{23} \end{gathered}$$

Define the total charge as below.

$$\begin{gathered} \mathrm{q}=\mathrm{Ne} \\ =\left(5.99\times {10}^{23}\right)\left(1.60\times {10}^{-19}\mathrm{C}\right) \\ =9.58\times {10}^4\mathrm{C} \end{gathered}$$

Determine the electrical force from equation (1) by substituting all known values.

$$\begin{gathered} F=\left(8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right){\left(\frac{9.58\times {10}^4\mathrm{C}}{3.84\times {10}^8\mathrm{m}}\right)}^2 \\ =560\mathrm{N} \\ =130\mathrm{lb} \end{gathered}$$

Hence, the force is repulsive in nature and this force is equal to the tension in the string.

The gravitational force expression is given as,

$F_g=G\frac{m_1{m}_2}{r^2}$

Here, $F_g$ is the gravitational force, $G$ is the gravitational constant, $r$is the distance between two masses $m_1$and $m_2$.

Here, $m_1=m_2$

Substitute all the values in the above equation,

$$\begin{gathered} F=\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\frac{{\left(1.0\times {10}^{-3}\mathrm{kg}\right)}^2}{{\left(3.84\times {10}^8\mathrm{m}\right)}^2} \\ =4.5\times {10}^{-34}\mathrm{N} \end{gathered}$$

Hence, the gravitational force is much less than the electrical force which makes it negligible in comparison.