Question

The charges of an electron and a positron are $\mathbf{\text{-eand+e}}$. The mass of each is$\mathbf{9}\mathbf{.11}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{}\mathbf{\text{\hspace{0.17em}kg}}$.What is the ratio of the electrical force to the gravitational force between an electron and a positron?

Short answer

The ratio of the electric force between the electron and the positron is $4.16\times {10}^{42}$.

Step-by-step solution

Given

The mass of electron and positron is $9.11\times {10}^{-31}$kg.

 Step 2: Determine the formulas for the electric forces:

Electric force between two charges$Q_1$and$Q_2$is given by$F_e=K\frac{Q_1{Q}_2}{r^2}$.

Gravitational force between two masses and is given by $F_g=\frac{G{m}_1{m}_2}{r^2}$.

Calculate the ratio of the electrical force to the gravitational force between an electron and a positron?

The electrical force between an electron and a positron separated by a distance r is${\mathbf{F}}_e\mathbf{=}\frac{{\mathbf{ke}}^2}{{\mathbf{r}}^2}$

On the other hand, the gravitational force between the two charges is$F_g=\frac{G{m}^2}{r^2}$. Thus, the ratio of the two forces is,

$\begin{array}{c}\frac{F_e}{F_g}=\frac{\frac{k{e}^2}{r^2}}{\frac{G{m}^2}{r^2}}\\ =\frac{k{e}^2}{G{m}^2}\end{array}$

Substitute the values and solve as:

$\begin{array}{c}\frac{F_e}{F_g}=\frac{(9\times {10}^9\frac{{\text{Nm}}^{\text{2}}}{{\text{C}}^{\text{2}}}){(1.60\times {10}^{-19}\text{\hspace{0.17em}C})}^2}{(6.67\times {10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{k{g}^2}){(9.11\times {10}^{-31}\text{\hspace{0.17em}kg})}^2}\\ =4.16\times {10}^{42}\end{array}$