91Ó°ÊÓ

The kinetic energy of a particle is given by

$\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$

Consider the couloumb’s law for the force.

$\mathbf{F}\mathbf{=}\mathbf{k}\frac{\left|q_1{q}_2\right|}{{\mathbf{r}}^{\mathbf{2}}}$

(a)

Consider the formula for the kinetic energy as follows:

$K=\frac{1}{2}m{v}^2$

Consider the charge on the proton is:

${\mathrm{q}}_{\mathrm{proton}}=1.602×{10}^{-19}\mathrm{C}$

Consider the mass of the proton is:

${\mathrm{m}}_{\mathrm{proton}}=1.6726×{10}^{-27} \mathrm{kg}$

Potential energy of a charged particle is as follows:

$U_1=k\frac{q{q}_0}{r}$

Substitute the values and solve as:

$U_1=k\frac{q{q}_0}{r}$

Substitute the values and solve as:

$$\begin{gathered} {\mathrm{U}}_1=\left(8.98755×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)×\frac{{\left(1.602×{10}^{-19}\mathrm{C}\right)}^2}{0.750×{10}^{-9}\mathrm{m}} \\ =3.076×{10}^{-19}\mathrm{J} \end{gathered}$$

Consider the expression for the kinetic energy in terms of the potential energy:

$$\begin{gathered} K_2=U_1 \\ =\frac{1}{2}m{v^2}_2+\frac{1}{2}m{v}_2^2 \end{gathered}$$

Therefore, speed of the particle will be

$$\begin{gathered} {\mathrm{v}}_2=\sqrt{\frac{{\mathrm{U}}_1}{{\mathrm{m}}_{\mathrm{proton}}}} \\ =\sqrt{\frac{3.076×{10}^{-19}\mathrm{J}}{1.6726×{10}^{-27}\mathrm{kg}}} \\ =1.356×{10}^4\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore, they will reach the maximum speed of $1.356×{10}^4\frac{\mathrm{m}}{\mathrm{s}}$.

(b)

For the distance to be minimum between two particles the acceleration is also maximum

Consider the expression for force:

$$\begin{gathered} F=ma \\ =\frac{q_1{q}_2}{r^{{}_2}} \end{gathered}$$

Consider the expression for acceleration:

$a=k\frac{q_1{q}_2}{m{r}^2}$

Substitute the values and solve as:

$$\begin{gathered} \mathrm{a}=\left(8.98755×{10}^9\frac{\mathrm{N}.{\mathrm{m}}^2}{{\mathrm{C}}^2}\right)×\frac{{\left(1.602×{10}^{-19}\mathrm{C}\right)}^2}{\left(1.6726×{10}^{-27}\mathrm{kg}\right){\left(0.750×{10}^{-9}\mathrm{m}\right)}^2} \\ =2.45×{10}^{17}\frac{\mathrm{m}}{{\mathrm{s}}^2} \end{gathered}$$

Therefore, they will reach the maximum acceleration of $2.45×{10}^{17}\frac{\mathrm{m}}{{\mathrm{s}}^2}$.