This question aims to find the **initial acceleration** of the **proton** released from a rest **point A** **5 mm** away.

The question is based on the concepts of **Coulomb’s Law. Coulomb’s law** is defined as the **electric force** between **two point charges** while they are at **rest** is called the **coulomb’s law.** The formula for **coulomb’s law** is given as:

\[ F = k \dfrac{ q_1 q_2 }{ r^2 } \]

## Expert Answer

The given information about the problem is:

\[ r = 5 mm \]

The **charge** on the all the **protons** in any **atom** is the same, which is given as:

\[ q = q_1 = q_2 = + 1.6 \times 10^ {-19} C \]

The **acceleration** of the **proton** is given by the **Newton’s second law** as:

\[ a = \dfrac{ F }{ m } \]

The **force F** is given by the **coulomb’s law** between **two protons** and the **mass** **m** of the **proton.** The formula for **force F** is given as:

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

\[ k = 9 \times 10^ {9} N m^2 C^ {-2} \]

\[ m = 1.67 \times 10^ {-27} kg \]

The equation becomes:

\[ a = \dfrac{ k q^2 }{ m r^2 } \]

Substituting the values, we get:

\[ a = \dfrac{ 9 \times 10^ {9} \times (1.6 \times 10^ {-19})^2 }{ 1.67 \times 10^ {-27} \times 0.005^2 } \]

Simplifying the equation, we get:

\[ a = 5.52 \times 10^ 3 m/s^2\ or 5.52 km /s^2 \]

## Numerical Result

The **initial acceleration** of the **proton** released from **rest position** is calculated to be:

\[ a = 5.52 \times 10^ 3 m/s^2 \]

## Example

In an experiment, a **proton** was **fixed** at a **position,** and **another proton** was released from a **position** **P** from a rest **3.5 mm** away. What will be the **initial acceleration** of the **proton** after the release?

The **distance** between **two protons** is given as:

** r = 3.5 mm**

The **total charge** on the **each proton** is **same** which is given as:

\[ q = q_1 = q_2 = + 1.6 \times 10^ {-19} C \]

We can use **Newton’s 2nd law,** where **force** **F** is given by **C****oulomb’s law** of **electrostatics.** The equation is given as:

\[ a = \dfrac{ F }{ m } \]

\[ F = \dfrac{ k q^2 }{ mr^2 } \]

Here:

\[ k = 9 \times 10^ {9} N m^2 C^ {-2} \]

\[ m = 1.67 \times 10^ {-27} kg \]

Substituting the values, we get:

\[ a = \dfrac{ 9 \times 10^ {9} \times (1.6 \times 10^ {-19})^2 }{ 1.67 \times 10^ {-27} \times 0.0035^2 } \]

\[ a = \dfrac{ 2.304 \times 10^ {-28} }{ 2.046 \times 10^ {-32} } \]

\[ a = 11262.4 m/s^2 \]

\[ a = 11.26 km /s^2 \]

The **initial acceleration** of the **proton** after it was released from rest is calculated to be **11.26 km per second squared.**