Question

Two protons are 2.0 fm apart.

a. What is the magnitude of the electric force on one proton due to the other proton?

b. What is the magnitude of the gravitational force on one proton due to the other proton?

c. What is the ratio of the electric force to the gravitational force.

Short answer

a. The magnitude of the electric force on one proton due to the other proton is $F_C=58N.$

b. The magnitude of the gravitational force on one proton due to the other proton is $F_G=4.7\times {10}^{-35}N$

c. The ratio of the electric force to the gravitational force is$\frac{F_C}{F_G}=1.25\times {10}^{36}$

Step-by-step solution

Content Introduction

The proton is a stable subatomic particle with a positive charge equal to that of an electron and a rest mass of $1,836$ times the mass of an electron.

Explanation (Part a)

Coulomb's Law describe this force as

$F_C=\frac{K{q}_1{q}_2}{r^2}$where K is the electrostatic constant and it is given by $9.0\times {10}^{9}N{m}^2/C^2$and the charge of each proton is $+1.6\times {10}^{-19}C$

$$\begin{gathered} F_C=\frac{K{q}_1{q}_2}{r^2} \\ {F}_C=\frac{(9.0\times {10}^{9}N{m}^2/C^2)(1.6\times {10}^{-19}C)(1.6\times {10}^{-19}C)}{{(2\times {10}^{-15}m)}^2} \\ {F}_C=58N \end{gathered}$$

Explanation (Part b)

Newton's law states the following:

$$\begin{gathered} F_G=G\frac{m_1{m}_2}{r^2} \\ {F}_G=(6.67\times {10}^{-11}N{m}^2/k{g}^2)\left(\frac{{(1.67\times {10}^{-27}kg)}^2}{{(2\times {10}^{-15}m)}^2}\right) \\ {F}_G=4.7\times {10}^{-35}N \end{gathered}$$

Explanation (Part c)

To calculate the ratio, divide the electric force $F_C$by the gravitational force $F_G$or the weight:

$$\begin{gathered} \frac{F_C}{F_G}=\frac{58N}{4.7\times {10}^{-35}N} \\ \frac{F_C}{F_G}=1.25\times {10}^{36}N \end{gathered}$$