Chapter 16, Problem 10P

To determine

The ratio of electric force and gravitational force.

Answer to Problem 10P

The ratio of electric force and gravitational force is $2.3\times {10}^{39}$ .

Explanation of Solution

Given:

Distance between electron and proton in an orbit, $r=0.53\times {10}^{-10}\text{m}$

Formula used:

Write the expression to calculate the magnitude of force acting between the electron and proton using Coulomb’s law.

  $F_e=k\frac{Q_1{Q}_2}{r^2}$.... (1)

Where,

The force acting between electron and proton is $F_e$ .

The Coulomb’s constant is $k$ and its value is $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^2$ .

The charge on electron and proton are $Q_1$ and $Q_2$ , respectively.

Write the expression to calculate the gravitational force acting between the electron and proton using gravitational force law.

  $F_g=G\frac{m_1{m}_2}{r^2}$.... (2) Where,

The gravitational force acting between electron and proton is $F_g$ .

The gravitational force constant is $G$ and its value is $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^2$ .

The mass of electron and proton are $m_1$ and $m_2$ , respectively.

Calculation:

Divide equation (1) by equation (2) to determine the ratio.

  $\begin{array}{c}\frac{F_e}{F_g}=\frac{k\frac{Q_1{Q}_2}{r^2}}{G\frac{m_1{m}_2}{r^2}}\\ =\frac{k{Q}_1{Q}_2}{G{m}_1{m}_2}\end{array}$

Substitute the values in the above expression.

  $\begin{array}{c}\frac{F_e}{F_g}=\frac{\left(8.988\times {10}^9\right)\left(1.6\times {10}^{-19}\right)\left(1.6\times {10}^{-19}\right)}{\left(6.67\times {10}^{-11}\right)\left(1.67\times {10}^{-27}\right)\left(9.1\times {10}^{-31}\right)}\\ \frac{F_e}{F_g}=2.3\times {10}^{39}\end{array}$

Conclusion:

Therefore, the ratio of electric force and gravitational force is $2.3\times {10}^{39}$ .