91Ó°ÊÓ

The charge ${\mathit{q}}_{\mathbf{1}}$and ${\mathit{q}}_{\mathbf{2}}$at any two instants of motion, should obey the conservation of mechanical energy, at the two particular positions a and b, expressed as

localid="1664264836838" $$\begin{gathered} T_a+U_a^2+k=T_b+U_b \\ \frac{1}{2}m{v^2}_a+k\frac{q_1{q}_2}{r_a}=\frac{1}{2}m{v}_b^2+k\frac{q_1{q}_2}{r_b} \end{gathered}$$

The two cases to be considered are, the case when $q_1$and $q_2$are at $0.400\mathrm{m}$separation when moving at a velocity, and when $q_1$and $q_2$are at $0.200\mathrm{m}$separation, when moving at an unknown velocity localid="1664263942211" $v_b$.

Rearranging the equation, and solve for$v_b$,

$$\begin{gathered} \frac{1}{2}m{v^2}_a-\frac{1}{2}m{v}_b^2=k\frac{q_1{q}_2}{r_b}-k\frac{q_1{q}_2}{r_a} \\ \frac{1}{2}m\left[v_b^2-v_b^2\right]=k{q}_1{q}_2\left[\frac{1}{r_b}-\frac{1}{r_a}\right] \\ \left[v_b^2-v_b^2\right]=\frac{2k{q}_1{q}_2}{m}\left[\frac{1}{r_b}-\frac{1}{r_a}\right] \\ {v}_b^{}=\sqrt{v_b^2-\frac{2k{q}_1{q}_2}{m}\left[\frac{1}{r_b}-\frac{1}{r_a}\right]} \end{gathered}$$$$\begin{gathered} \frac{1}{2}\mathrm{m}{{\mathrm{v}}^2}_{\mathrm{a}}-\frac{1}{2}{\mathrm{mv}}_{\mathrm{b}}^2=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{b}}}-\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{a}}} \\ \frac{1}{2}\mathrm{m}\left[{{\mathrm{v}}^2}_{\mathrm{a}}-{{\mathrm{v}}^2}_{\mathrm{a}}\right]=\mathrm{k}={\mathrm{q}}_1{\mathrm{q}}_2\left[\frac{1}{{\mathrm{r}}_{\mathrm{a}}}-\frac{1}{{\mathrm{r}}_{\mathrm{a}}}\right] \\ \left[{{\mathrm{v}}^2}_{\mathrm{a}}-{{\mathrm{v}}^2}_{\mathrm{a}}\right]=\frac{2\mathrm{k}.{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{m}}\left[\frac{1}{{\mathrm{r}}_{\mathrm{a}}}-\frac{1}{{\mathrm{r}}_{\mathrm{a}}}\right] \\ {{\mathrm{v}}^{}}_{\mathrm{a}}=\sqrt{{{\mathrm{v}}^2}_{\mathrm{a}}-\frac{2\mathrm{k}.{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{m}}\left[\frac{1}{{\mathrm{r}}_{\mathrm{a}}}-\frac{1}{{\mathrm{r}}_{\mathrm{a}}}\right]} \end{gathered}$$

he speed when the charges are at a separation of $0.200\mathrm{m}$is obtained by substituting the values $v_a=800m/s$, $m=4.00*{10}^{-3}kg$, $r_b=0.200m$, $q_1=+5.00*{10}^{-4}C$, $q_2=-3.00*{10}^{-4}C$in $v_b^{}=\sqrt{v_b^2-\frac{2k{q}_1{q}_2}{m}\left[\frac{1}{r_b}-\frac{1}{r_a}\right]}$

Hence, the speed when the charges reached $0.200\mathrm{m}$distance of separation is $1.52km/s$.