# In an experiment in space, one proton is fixed and other is released from rest (point A), from a distance of 5 mm away. What is the initial acceleration of the proton after it is released?

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:

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

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 Coulomb’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.