Question

A point charge $q_1=9.1\mu \text{C}$ is held fixed at origin. A second point charge $q_2=-0.42\mu \text{C}$ and a mass $3.2\times {10}^{-4}\text{kg}$ is placed on the $\text{x-axis}$, $0.96\text{m}$ from the origin. The second point charge is released at rest. What is its speed when it is $0.24\text{m}$ from the origin?

• A. $15{\text{ms}}^{-1}$
• B. $10{\text{ms}}^{-1}$
• C. $42{\text{ms}}^{-1}$
• D. $26{\text{ms}}^{-1}$

Answer 1

The correct option is D $26{\text{ms}}^{-1}$

Given,
$q_1=9.1\mu \text{C}=9.1\times {10}^{-6}\text{C}$

$q_2=-0.42\mu \text{C}=-0.42\times {10}^{-6}\text{C}$

mass of second point charge, $m=3.2\times {10}^{-4}\text{kg}$

initial distance of second point charge from origin, $r_i=0.96\text{m}$

final distance of second point charge from origin, $r_f=0.24\text{m}$

Now,
Initial energy of system$=\frac{q_1{q}_2}{4\pi {\epsilon }_0}\frac{1}{r_i}$

When $q_2$ reached at $r_f=0.24\text{m}$,

energy of system= $\frac{1}{2}m{v}^2+\frac{q_1{q}_2}{4\pi {\epsilon }_0}\frac{1}{r_f}$


From conservation of energy, we have

$\text{Initial energy of the system=Final energy of the system}$

$\frac{q_1{q}_2}{4\pi {\epsilon }_0}\frac{1}{r_i}=\frac{1}{2}m{v}^2+\frac{q_1{q}_2}{4\pi {\epsilon }_0}\frac{1}{r_f}$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{q_1{q}_2}{4\pi {\epsilon }_0}(\frac{1}{r_i}-\frac{1}{r_f})$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{q_1{q}_2}{4\pi {\epsilon }_0}\left(\frac{r_f-r_i}{r_i{r}_f}\right)$

$\Rightarrow v=\sqrt{\frac{q_1{q}_2}{2\pi {\epsilon }_{0}m}\left(\frac{r_f-r_i}{r_i{r}_f}\right)}$

Substituting the values in the above equation,

$\Rightarrow v=\sqrt{\frac{(9.1\times {10}^{-6})(-0.42\times {10}^{-6})\times 2\times 9\times {10}^9}{3.2\times {10}^{-4}}\left(\frac{0.24-0.96}{\left(0.24\right)\left(0.96\right)}\right)}(\because \frac{1}{4\pi {\epsilon }_0}=9\times {10}^9)$

$\therefore v=26{\text{ms}}^{-1}$

Hence, option (b) is correct answer.