23.46: A point charge${\mathbf{q}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mC}$is held fixed in space. From a horizontal distance of$\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{cm}$, a small sphere with mass$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{kg}$and charge${\mathbf{q}}_{\mathbf{2}}\mathbf{=}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{mC}$is fired toward the fixed charge with an initial speed of. Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is$\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$?

The attractive electrostatic force experienced by two charged bodies each having charges ${\mathbf{q}}_{\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{q}}_{\mathbf{2}}$is $\mathit{F}\mathbf{=}\mathbf{-}\mathit{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{d}}^{\mathbf{2}}}$, (negative sign for oppositely charged attracting charges)where k is the constant $\mathit{k}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{p}{\mathbf{e}}_{\mathbf{0}}}\mathbf{=}\mathbf{9}\mathbf{*}{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathit{N}{\mathit{m}}^{\mathbf{2}}\mathbf{/}{\mathit{C}}^{\mathbf{2}}$, d is the distance of separation between the charges, at the instance, the force acts.

The force is also proportional to acceleration, as per the Newton’s law, $\mathbf{F}\mathbf{=}\mathbf{ma}$. Combining, $\mathit{m}\mathit{a}\mathbf{=}\mathit{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{d}}^{\mathbf{2}}}$, then the acceleration is $\mathit{a}\mathbf{=}\mathit{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{m}{\mathbf{d}}^{\mathbf{2}}}$. At the final instance of consideration, the distance is also represented as ${\mathit{r}}_{\mathbf{f}}$. Then,$\mathit{a}\mathbf{=}\mathit{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{m}{{\mathbf{r}}_{\mathbf{f}}}^{\mathbf{2}}}$……(1)

Given are the initial and final velocities, which can be related to each other by energy conservation law, ${\mathit{U}}_{\mathbf{i}}\mathbf{+}{\mathit{T}}_{\mathbf{i}}\mathbf{=}{\mathit{U}}_{\mathbf{f}}\mathbf{+}{\mathit{T}}_{\mathbf{f}}$, where ${\mathit{U}}_{\mathbf{i}}$ is the initial potential energy (${\mathit{U}}_{\mathbf{i}}\mathbf{=}\mathit{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}_{\mathbf{i}}}$), ${\mathit{U}}_{\mathbf{f}}$is the final potential energy (${\mathit{U}}_{\mathbf{i}}\mathbf{=}\mathit{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}_{\mathbf{f}}}$), ${\mathit{T}}_{\mathbf{i}}$is the initial kinetic energy localid="1665139253434" $\mathbf{(}{\mathit{T}}_{\mathbf{i}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{{\mathit{v}}_{\mathbf{i}}}^{\mathbf{2}}\mathbf{)}$,${\mathit{T}}_{\mathbf{f}}$is the final kinetic energy.

Then, the energy conservation law becomes$\mathit{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}_{\mathbf{i}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{i}}^{\mathbf{2}}\mathbf{=}\mathit{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}_{\mathbf{f}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{f}}^{\mathbf{2}}$.

Solve for ${\mathit{r}}_{\mathbf{f}}$as $\mathit{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}_{\mathbf{i}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{i}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{f}}^{\mathbf{2}}\mathbf{=}\mathit{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}_{\mathbf{f}}}$; then ${\mathit{r}}_{\mathbf{f}}\mathbf{=}\mathit{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{k}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}_{\mathbf{i}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{\mathbf{v}}_{\mathbf{i}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{\mathbf{v}}_{\mathbf{f}}^{\mathbf{2}}}$…..(2)

Substituting the values of $k=9*{10}^{9}N{m}^2/C^2$, $q_1=5\mu C=5*{10}^{-6}C$, $q_2=2\mu C=2*{10}^{-6}C$, $r_i=6cm=0.06m$, $\mathrm{m}=4.00*{10}^{-3}\mathrm{kg}$, $v_i=40.0m/s$, $v_f=25.0m/s$,calculate $r_f$.

$$\begin{gathered} r_f=9*{10}^9*\frac{(5*{10}^{-6}*2*{10}^{-6})}{\left(9*{10}^9\right){\displaystyle \frac{(5*{10}^{-6}*2*{10}^{-6})}{0.06}}+\left({\displaystyle \frac{1}{2}}*4*{10}^{-3}*{40}^2\right)-\left(\frac{1}{2}*4*{10}^{-3}*{25}^2\right)} \\ {r}_f=9*{10}^9*\frac{\left({10}^{-11}\right)}{{\displaystyle \frac{\left(9*{10}^{-2}\right)}{0.06}}+\left(3.2\right)-\left(1.25\right)} \\ {r}_f=\frac{\left(9*{10}^{-2}\right)}{1.5+\left(3.2\right)-\left(1.25\right)} \\ {r}_f=0.026m \end{gathered}$$

Thus, the distance at which the velocity is $\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$is $r_f=0.026m$.

The acceleration at the distance where the velocity has reduced to $25\mathrm{m}/\mathrm{s}$ is to be found by substituting the values of$k=9*{10}^{9}N{m}^2/C^2$, $q_1=5\mu C=5*{10}^{-6}C$, $q_2=2\mu C=2*{10}^{-6}C$,

localid="1665141539833" $m=4.00*{10}^{-3}\mathrm{kg}$, $r_f=0.026m$in $a=k\frac{q_1{q}_2}{m{r_f}^2}$.

$$\begin{gathered} a=\left(9*{10}^9\right)\frac{\left({10}^{-11}\right)}{\left[2.704*{10}^{-6}\right]} \\ a=33284.02 \\ a=3.3*{10}^{4}m/s^2 \end{gathered}$$

Thus, at the instant, the velocity is $\mathbf{25}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, the acceleration of the approaching body is $3.3*{10}^{4}m/s^2$.