Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated protons separated by\({\bf{2}}.{\bf{00}}{\rm{ }}{\bf{nm}}\)(a typical distance between gas atoms). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.

According to the Newton’s second law of motion, the rate of change of momentum is directly proportional to the force applied.

\(\begin{aligned} {\rm{F\mu }}\frac{{{\rm{\Delta p}}}}{{{\rm{\Delta t}}}}\\{\rm{ = k}}\left( {\frac{{{\rm{mv - mu}}}}{{{\rm{\Delta t}}}}} \right)\\{\rm{ = k}}\left( {{\rm{m}}\left( {\frac{{{\rm{v - u}}}}{{{\rm{\Delta t}}}}} \right)} \right)\end{aligned}\)

Here, F is the force, \(\Delta p\) is the change in momentum, \({\rm{\Delta t}}\) is the change in time, k is the proportionality constant (k =1), m is the mass, v is the final velocity, and u is the initial velocity.

The acceleration of the body is,

\(a\, = \,\frac{{v{\rm{ }} - {\rm{ }}u}}{{\Delta t}}\)

From equation (1.1) and (1.2),

\(F = ma\)

The electrostatic force between two protons is,

\(F = \frac{{Kqq}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant \(\left( {K = 9 \times {{10}^9}{\rm{ N \times }}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), q is the charge on proton \(\left( {q = 1.6 \times {{10}^{ - 19}}{\rm{ }}C} \right)\), and r is the separation between the protons \(\left( {r = 2.00{\rm{ }}nm} \right)\).

From equation (1.3), and (1.4),

\(ma = \frac{{Kqq}}{{{r^2}}}\)

Here, m is the mass of proton \(\left( {m = 1.67 \times {{10}^{ - 27}}{\rm{ }}kg} \right)\), and a is the acceleration of proton.

The expression for the acceleration of proton is,

\(a = \frac{{Kqq}}{{m{r^2}}}\)

Substituting all known values,

\(\begin{aligned} a &= \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right) \times \left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right) \times {{\left( {2.00{\rm{ nm}}} \right)}^2}}}\\ &= \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right) \times \left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right) \times {{\left( {\left( {2.00{\rm{ nm}}} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1{\rm{ nm}}}}} \right)} \right)}^2}}}\\ &= 3.45 \times {10^{16}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\end{aligned}\)

Hence, the acceleration of the proton is \(3.45 \times {10^{16}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\).