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According to law of conservation of energy

$$\begin{gathered} \mathbf{螖碍贰}\mathbf{-}\mathbf{螖鲍}\mathbf{=}\mathbf{0} \\ {\mathbf{K}}_{\mathbf{a}}\mathbf{+}{\mathbf{U}}_{\mathbf{a}}\mathbf{=}{\mathbf{K}}_{\mathbf{b}}\mathbf{+}{\mathbf{U}}_{\mathbf{b}} \end{gathered}$$

Where Ua=0

${\mathrm{K}}_{\mathrm{a}}=\frac{1}{{\mathrm{r}}_{\mathrm{i}}}{{\mathrm{mv}}_{\mathrm{b}}}^2+\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}}$

According to law of conservation of momentum

localid="1664883029399" ${\mathbf{P}}_{\mathbf{a}}\mathbf{=}{\mathbf{P}}_{\mathbf{b}}$

$$\begin{gathered} {\mathrm{mv}}_{\mathrm{a}}\mathrm{b}={\mathrm{mv}}_{\mathrm{b}}{\mathrm{r}}_{\mathrm{i}} \\ \frac{{\mathrm{v}}_{\mathrm{a}}{\mathrm{b}}_{\mathrm{i}}}{{\mathrm{r}}_{\mathrm{i}}}={\mathrm{v}}_{\mathrm{b}} \end{gathered}$$

Substitute to get Ka

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{1}{2}{{\mathrm{mv}}^2}_{\mathrm{b}}+\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}} \\ =\frac{1}{2}\mathrm{m}{\left(\frac{{\mathrm{v}}_{\mathrm{a}}\mathrm{b}}{{\mathrm{r}}_{\mathrm{i}}}\right)}^2+\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}} \\ =\frac{1}{2}{{\mathrm{mv}}^2}_{\mathrm{b}}{\left(\frac{\mathrm{b}}{{\mathrm{r}}_{\mathrm{i}}}\right)}^2+\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}} \\ ={\mathrm{K}}_{\mathrm{a}}{\left(\frac{\mathrm{b}}{{\mathrm{r}}_{\mathrm{i}}}\right)}^2+\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}} \\ {\mathrm{K}}_{\mathrm{a}}{{\mathrm{r}}_{\mathrm{i}}}^2-{\mathrm{kq}}_1{\mathrm{q}}_2{\mathrm{r}}_{\mathrm{i}}-{\mathrm{K}}_{\mathrm{a}}{\mathrm{b}}^2\mathrm{i}=0 \end{gathered}$$

Substitute to get value of r1at b1:

$[1.76脳{10}^{-12}]{{\mathrm{r}}_1}^2-3.78脳{10}^{-26}{\mathrm{r}}_1-1.76脳{10}^{-12}{(1.00脳{10}^{-12})}^2=0$

Solve the above equation to get r1:

$$\begin{gathered} {\mathrm{r}}_1=\frac{-\mathrm{b}卤\sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ =\frac{-(-3.78脳{10}^{-26})卤\sqrt{\left(-3.78脳{10}^{-26}\right)-4\left[1.76脳{10}^{-12}\right]}\left[1.76脳{10}^{-12}\left(1.00脳{10}^{-12}\right)\right]}{2\left[1.76脳{10}^{-12}\right]} \\ =1.01脳{10}^{-12}\mathrm{m} \end{gathered}$$

Therefore, the distance of closest approach when b = 1.00 x 10-12 m is r1=1.01 x 10-12 m.

Substitute to get value of r2at b2:

$[1.76脳{10}^{-12}]{{\mathrm{r}}_1}^2-3.78脳{10}^{-26}{\mathrm{r}}_1-1.76脳{10}^{-12}{(1.00脳{10}^{-13})}^2=0$

Solve the above equation to get r2:

$$\begin{gathered} \mathrm{r}2=\frac{-\mathrm{b}卤\sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ =\frac{-(-3.78脳{10}^{-26})卤\sqrt{\left(-3.78脳{10}^{-26}\right)-4\left[1.76脳{10}^{-12}\right]}\left[1.76脳{10}^{-12}\left(1.00脳{10}^{-12}\right)\right]}{2\left[1.76脳{10}^{-12}\right]} \\ =1.11脳{10}^{-13}\mathrm{m} \end{gathered}$$

Therefore, the distance of closest approach when b = 1.00 x 10-13 m is r2=1.11 x 10-13 m.

Substitute to get value of r3at b3:

$[1.76脳{10}^{-12}]{{\mathrm{r}}_1}^2-3.78脳{10}^{-26}{\mathrm{r}}_1-1.76脳{10}^{-12}{(1.00脳{10}^{-13})}^2=0$

Solve the above equation to get r3:

$$\begin{gathered} {\mathrm{r}}_3=\frac{-\mathrm{b}卤\sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ =\frac{-(-3.78脳{10}^{-26})卤\sqrt{\left(-3.78脳{10}^{-26}\right)-4\left[1.76脳{10}^{-12}\right]}\left[1.76脳{10}^{-12}\left(1.00脳{10}^{-12}\right)\right]}{2\left[1.76脳{10}^{-12}\right]} \\ =2.54脳{10}^{-14}\mathrm{m} \end{gathered}$$

Therefore, the distance of closest approach when b = 1.00 x 10-14 m is r3=2.54 x 10-14 m.