This problem aims to brief the concepts of attractive and repulsive forces between two point charges having the same magnitudes. This problem requires the knowledge of **field forces, Coulomb’s law, **and **the Law of conservation of energy**, which is briefly explained in the solution below.

## Expert Answer

**Coulomb’s law** states that the maximum force between the two charges having the magnitudes $q1$ and $q2$ and the distance $r$ is equal to:

\[ F = \dfrac{1}{4 \pi \epsilon_o} \dfrac{|q_1 q_2|}{r^2} \]

Here, $ \dfrac{1}{4 \pi \epsilon_o} $ is known as the **Coulomb’s constant** and is denoted by $k$ or $k_e$, where its value always remains constant and is given by $ 9.0 \times 10^9 N. m^2/C^2 $.

On the other hand, $q1$ and $q2$ are two equally charged protons, and their charge is equal to $1.602 \times 10^{-19} C$

$r$ is the distance at which the protons exert the maximum electric force on each other.

According to the **Law of Conservation of Energy**, proton initial **K.E.** is equal to its final **P.E.**, therefore, we can write something like this:

\[KE_{Initial} = PE_{Final}\]

\[\dfrac{1}{2} mv^2=k \dfrac{e^2}{r}\]

Since $r$ is the unknown here, the equation becomes:

\[r=\dfrac{2ke^2}{mv^2}\]

Here, $m$ is the mass of one proton and is given as $ 1.67 \times 10^-27 kg.$.

Solving the equation for $r$ by substituting the values back in:

\[r=\dfrac{( 9.0 \times 10^9) (1.602\times 10^{-19})^2}{(1.67\times 10^-27)(3.50 \times 10^5) ^2}\]

\[r=1.127 \times 10^{-12}\]

As $r$ is the minimum distance at which the two protons exert maximum force on each other, so the maximum electrostatic force $F$ can be found by plugging in the value of $k$, $e$, and $r$:

\[F=k\dfrac{e^2}{r^2}\]

## Numerical Answer

\[F=9.0\times 10^9 \dfrac{(1.602 \times 10^{-19})^2}{r^2}\]

\[F=0.000181 N\]

The maximum electric force these protons will exert on each other while keeping a minimum distance between them is $0.000181 N$.

## Example

Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of $2.30 \times 10^5 m/s$, measured relative to the earth. Find the maximum electrical force that these protons will exert on each other.

As our first step, we will find the $r$ at which these protons will exert the maximum force. Here, the value of $r$ can be easily computed by referring to **Law of conservation of energy**, in which initial **Kinetic Energy** equals to the final **Potential Energy**. It is expressed as:

\[r=\dfrac{ke^2}{mv^2}\]

\[r = \dfrac{( 9.0 \times 10^9) (1.602 \times 10^{-19}) ^2}{(1.67 \times 10^-27)(2.30 \times 10^5) ^2} \]

\[ r = 2.613 \times 10^{-12}\]

After calculating $r$, step $2$ is to calculate electric force $F$ at the obtained $r$, and the expression for $F$ is given as:

\[ F = k \dfrac{e^2}{r^2} \]

\[ F = 9.0 \times 10^9 \dfrac{(1.602 \times 10^{-19})^2}{r^2} \]

\[ F = 3.3817 \times 10^{-5} N \]

Note that if the value of $e$ (which is the product of the charge quantity of the protons) is positive, the electrostatic force between the two charges is repulsive. If it is negative, the force between them should be attractive.