Question

A positive point charge${\mathbf{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{C}$is held at a fixed position. A small object with mass$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{*}\mathbf{}\mathbf{103}\mathbf{}\mathbf{kg}$and charge${\mathit{q}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{C}$is projected directly at${\mathbf{q}}_{\mathbf{1}}$. Ignore gravity. When${\mathbf{q}}_{\mathbf{2}}$is$\mathbf{0}\mathbf{.}\mathbf{400}\mathbf{}\mathbf{m}$away, its speed is$\mathbf{800}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. What is its speed when it is$\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{}\mathbf{m}$from$\mathit{q}\mathbf{1}$?

Short answer

The speed of $q_2$when it is $0.200\mathrm{m}$from $q_1$is $1.52km/s$.

Step-by-step solution

Step-1: The conservation law

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$v_b$,

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

Step-2: Calculating the speed

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 $v_b^{}=\sqrt{v_b^2-\frac{2k{q}_1{q}_2}{m}\left[\frac{1}{r_b}-\frac{1}{r_a}\right]}$

localid="1664265627279" $$\begin{gathered} v_b=\sqrt{\left[{800}^2-\left\{\frac{\left[2*9*{10}^9*"5.00*{10}^{-4}\right]}{4.00*{10}^{-3}kg}[\frac{1}{0.20}-\frac{1}{0.40}\right\}\right]} \\ {v}_b=\sqrt{\left[{800}^2-\left\{\frac{\left[2*9*{10}^9*"5.00*{10}^{"}-4*-3*{10}^{-4}\right]}{4.00*{10}^{-3}kg}[\frac{1}{0.20}-\frac{1}{0.40}\right\}\right]} \\ {v}_b=1525.6146m/s \\ {v}_b=1.52km/s \end{gathered}$$

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