Q153 / 200 · AP EAMCET 2001 · Physics
Two electric charges of $9 \mu \mathrm{C}$ and $-3 \mu \mathrm{C}$ are placed $0.16 \mathrm{~m}$ apart in air. There will be a point ' $P$ ', at which electric potential is zero on the line joining two charges and in between them. The distance of $P$ from $9 \mu \mathrm{C}$ is
- A$0.14 \mathrm{~m}$
- Correct answerB$0.12 \mathrm{~m}$
- C$0.08 \mathrm{~m}$
- D$0.06 \mathrm{~m}$
Solution
$q_1=9 \mu \mathrm{C}, q_x=-3 \mu \mathrm{C}, r=0.16 \mathrm{~m}$
Potential at point ' $P$ ', $V_1+V_3=0$ or $V_1=-V_2$ $\begin{aligned} \frac{1}{4 \pi x_1} \frac{q_1}{r_1} & =\frac{-1}{4 \pi x_0} \frac{q_3}{r_2} \\ \frac{9}{x} & =\frac{-(-3)}{(0.16-x)}\end{aligned}$ $\Rightarrow \quad \begin{aligned} 3(0.16-x) & =x \\ 0.48-3 x & =x \\ 4 x=0.48 \Rightarrow x & =0.12 \mathrm{~m}\end{aligned}$
Potential at point ' $P$ ', $V_1+V_3=0$ or $V_1=-V_2$ $\begin{aligned} \frac{1}{4 \pi x_1} \frac{q_1}{r_1} & =\frac{-1}{4 \pi x_0} \frac{q_3}{r_2} \\ \frac{9}{x} & =\frac{-(-3)}{(0.16-x)}\end{aligned}$ $\Rightarrow \quad \begin{aligned} 3(0.16-x) & =x \\ 0.48-3 x & =x \\ 4 x=0.48 \Rightarrow x & =0.12 \mathrm{~m}\end{aligned}$
Hence, the correct answer is option: 'B'.