Two point charges $-10 \mu \mathrm{C}$ and $+5 \mu \mathrm{C}$ are situated on $X$-axis at $x=0$ and $x=\sqrt{2} \mathrm{~m}$. The point along the $X$-axis where the electric field becomes zero is

• A

  $x=(\sqrt{2}-1) \mathrm{m}$
• B

  $x=2(\sqrt{2}-1) \mathrm{m}$
• Correct answer
  C

  $\quad x=2(\sqrt{2}+1) \mathrm{m}$
• D

  $x=(\sqrt{2}+1) \mathrm{m}$

Solution

$\begin{aligned} & \text { (c) } \mathrm{q}_1=-10 \mu \mathrm{C}, \mathrm{q}_2=5 \mu \mathrm{C} \\ & \mathrm{r}=\sqrt{2} \mathrm{~m}\end{aligned}$
The point on $x$-axis where electric field is zero, $\begin{aligned} & x=\left(\frac{\sqrt{q_2}}{\sqrt{q_1}-\sqrt{q_2}}\right) r=\left(\frac{\sqrt{5}}{\sqrt{10}-\sqrt{5}}\right) \sqrt{2} \\ & =\frac{\sqrt{2}}{\sqrt{2}-1} \mathrm{~m} \end{aligned}$ $\therefore$ Distance of neutral point $\mathbf{P}$, $\mathrm{d}=\sqrt{2}+\mathrm{x}=\sqrt{2}+\frac{\sqrt{2}}{\sqrt{2}-1}=2(\sqrt{2}+1)_{\mathrm{m}}$

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