Two fixed charges A and B of $5 \mu \mathrm{C}$ each are separated by a distance of $6 \mathrm{~m}$. $\mathrm{C}$ is the mid point of the line joining $A$ and $B$. A charge $Q$ of $-5 \mu \mathrm{C}$ is shot perpendicular to the line joining A and $B$ through $C$ with a kinetic energy of $0.06 \mathrm{~J}$. The charge $\mathrm{Q}$ comes to rest at a point D. The distance $C D$ is

• Correct answer
  A

  $4 \mathrm{~m}$
• B

  $3 \mathrm{~m}$
• C

  $\sqrt{3} \mathrm{~m}$
• D

  $3 \sqrt{3} \mathrm{~m}$

Solution

By conservation of energy Loss of PE = Gain in KE $\frac{2 \mathrm{a}^{2}}{4 \pi \varepsilon_{0}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\sqrt{\mathrm{r}^{2}+\mathrm{x}^{2}}}\right)=\mathrm{K}$

or $\begin{aligned} 2 \times 9 \times 10^{9} \times &\left(5 \times 10^{-6}\right)^{2}\left(\frac{1}{3}-\frac{1}{\sqrt{3^{2}+x^{2}}}\right) \\ &=0.06 \end{aligned}$ or $\quad \frac{1}{3}-\frac{1}{\sqrt{9+x^{2}}}=\frac{2}{15}$ or $$ x=4 m $$

Hence, the correct answer is option: 'A'.