Question

An electron and a positron are separated by distance r. Findthe ratio of the gravitational force to the electric force between them. From the result, what can you conclude concerning the forces acting between particles detected in a bubble chamber?

(Should gravitational interactions be considered?)

Short answer

The ratio between the two forces is$2.4\times {10}^{-43}$.

Since,$F_G\ll F_e$,we can neglect the gravitational force acting between particles in a bubble chamber.

Step-by-step solution

Identifying the data given in the question.

The distance between the electron and a positron is r.

We know,

The charge on the electron and positron is

$Q_e=Q_p=e=1.6\times {10}^{-19} \mathrm{C}$

The mass of the electron and positron is

$m_e=m_p=m=9.11\times {10}^{-31} \mathrm{kg}$

Concept used to solve the question.

When two charged particles are placed at some distance then the electric forces and gravitational force acts between them

• The Gravitational Force can be given as

${\mathit{F}}_{\mathbf{G}}\mathbf{=}\frac{\mathbf{G}{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$

Where m₁and m₂ represent the mass, r represents the distance and G represents the gravitational constant.

• The Electric Force between can be given as

${\mathit{F}}_{\mathbf{e}}\mathbf{=}\frac{\mathbf{k}{\mathbf{Q}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$

Where Q₁and Q₂represent the charges, r represents the distance and k represents the proportionality constant.

Find the ratio of two forces.

The gravitational force between electron and positron be given as

$F_G=\frac{G{m}^2}{r^2}$

The Electric force between them can be given as,

$F_e=\frac{k{e}^2}{r^2}$

The ratio of two forces

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

We know,

G = 6.67 x 10^-11 N.kg²/m², k=9.0 x 10⁹N.m²/c²

Substituting the values,

$\begin{array}{l}\frac{F_G}{F_e}=\frac{6.67\times {10}^{-11}{\text{\hspace{0.17em}N.kg}}^{\text{-2}}/{\text{m}}^{\text{2}}{(9.11\times {10}^{-31}\text{\hspace{0.17em}\hspace{0.17em}kg})}^2}{9.0\times {10}^9{\text{N.m}}^{\text{2}}/{\text{C}}^{\text{2}}{(1.6\times {10}^{-19}\text{\hspace{0.17em}\hspace{0.17em}C})}^2}\\ \frac{F_G}{F_e}=2.4\times {10}^{-43}\end{array}$

Therefore, the ratio between the two forces is$2.4\times {10}^{-43}$

Since, $F_G\ll F_e$, we can neglect the gravitational force acting between particles in a bubble chamber.