Chapter 20, Problem 8STP

To determine

To Choose: The correct option.

Answer to Problem 8STP

D. $3.1\times {10}^{35}$

Explanation of Solution

Given:

The mass of the alpha particle, $m=6.68\times {10}^{-27}\text{kg}$

The charge of the alpha particle, $q=3.2\times {10}^{-19}\text{C}$

Formula used:

The electrostatic force between two charges, $q_1$ and $q_2$ , $r$ distance apart is given by,

  $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

Where ${\epsilon }_0=\text{ permittivity of free space}=8.85\times {10}^{-12}\text{F/m}$

And, Law of Universal Gravitation,

  $F=\frac{G{m}_1{m}_2}{r^2}$

Where $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the distance between the masses.

It is known that the value of gravitational constant is,

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Calculation:

Let the distance between the alpha particles be $r$ .

Then, the electrostatic force between the two alpha particles is,

  $F_E=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$

And, the gravitational force between the particles is,

  $F_G=\frac{G{m}^2}{r^2}$

Then, the ratio of the electrostatic force to the gravitational force between two alpha particles will be,

  $\begin{array}{c}\frac{F_E}{F_G}=\frac{\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}\right)}{\left(\frac{G{m}^2}{r^2}\right)}\\ \frac{F_E}{F_G}=\frac{\left(q^2\right)}{\left(4\pi {\epsilon }_{0}G{m}^2\right)}\\ =\frac{{\left(3.2\times {10}^{-19}\right)}^2}{\left(4\pi \times 8.85\times {10}^{-12}\times 6.67\times {10}^{-11}\times {\left(6.68\times {10}^{-27}\right)}^2\right)}\\ =3.1\times {10}^{35}\end{array}$

Conclusion:

Hence, the ratio of the electrostatic force to the gravitational force between two alpha particles is $3.1\times {10}^{35}$ .