Use data from the inside back cover to calculate the gravitational and electric forces two electrons exert on each other when they are$\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{m}$apart (about one atomic radius). Which interactions between two electrons is stronger, the gravitational attraction or the electric repulsion? If the two electrons are at rest, will they begin to move toward each other or away from each other? Note that since both the gravitational and the electric forces depend on the inverse square distance, this comparison holds true at all distances, not just at a distance of$\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{m}$.

The given data can be listed below as,

• The distance between the electrons is$1\times {10}^{{}^{-}10}\hspace{0.33em}\mathrm{m}$

The gravitational law states that the force exerted by the particle is directly proportional to the product of the masses and inversely proportional to the square of the distances amongst them.

The Coulomb’s law states that the force exerted by the particle is directly proportional to the product of the charges and inversely proportional to the square of their distances.

The equation of the gravitational and the electrostatic force gives the gravitational and the electrostatic force of the electrons.

From Newton’s gravitational law, the gravitational force exerted by the electrons can be expressed as:

$F=\frac{G{m}_1{m}_2}{r^2}$

Here, F is the gravitational force; G is the gravitational constant, $M_{1}and{m}_2$are the mass of the electrons that is $9.1\times {10}^{-31}kg$and r is the distance amongst them.

Substituting the values in the above equation, we get-

$$\begin{gathered} \mathrm{F}=\frac{\left(6.67\times {10}^{-1}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\times {\left(9.1\times {10}^{-31}\mathrm{kg}\right)}^2}{{\left(1\times {10}^{-10}\mathrm{m}\right)}^2} \\ \mathrm{F}=5.523\times {10}^{-51}\mathrm{N} \end{gathered}$$

According to the coulomb’s law, the magnitude of the electric force can be expressed as:

$F=k\frac{q_1{q}_2}{r^2}$

Here, F is the magnitude of the electric force; k is the coulomb’s constant that is about$8.99\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2,q_1\text{and}{q}_2$the charges of the electrons that is$-1.6\times {10}^{-19}\hspace{0.33em}C$and r is the distance amongst them.

Substituting the values in the above equation, we get-

$$\begin{gathered} \mathrm{F}=8.99\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{{\left(-1.6\times {10}^{-19}\mathrm{C}\right)}^2}{{\left(1\times {10}^{-10}\mathrm{m}\right)}^2} \\ \mathrm{F}=2.301\times {10}^{-8}\mathrm{N} \end{gathered}$$

Thus, the gravitational and the electric forces two electrons exert on each other are$5.523\times {10}^{-51}\mathrm{N}\text{and}2.301\times {10}^{-8}\mathrm{N}$ respectively. The electric repulsion between the two electrons is stronger.

As the electrons has a negative charge, so they will repel each other.

Thus, the electrons will move away from each other.