Q144 / 160 · TS EAMCET 2023 (13 May Shift 2) · Physics
Two electric charges $+2 \mu \mathrm{C}$ and $-4 \mu \mathrm{C}$ are separated by a distance $3 \mathrm{~m}$ in air. At a point $\mathrm{P}$ located on the line joining the two charges and in between them, the electric potential is zero. Then the electric field at a point $\mathrm{P}\left(\right.$ in $\left.\mathrm{NC}^{-1}\right)$ is
- A9,000
- Correct answerB18,000
- C12,000
- D27,000
Solution
$\mathrm{Q}_1=+2 \times 10^{-6} \mathrm{C} ; \mathrm{Q}_2=-4 \times 10^{-6} \mathrm{C}$ Distance, $\mathrm{d}=3 \mathrm{~m}$
Electric potential, $V=\frac{2 \times 10^{-6}}{4 \pi \epsilon_0(3-x)}+\frac{(-4) \times 10^{-6}}{4 \pi \epsilon_0 x}$ $$ \begin{aligned} & 0=\frac{2}{3-x}-\frac{4}{x} \\ & \Rightarrow 0=\frac{1}{3-x}-\frac{2}{x} \Rightarrow \frac{1-}{3-x}=\frac{2}{x} \\ & \Rightarrow x=6-2 x \\ & \Rightarrow 3 x=6 ; x=2 m \end{aligned} $$ Electric field at a point $P$ is $$ \begin{aligned} & E=\frac{Q_2}{4 \pi \epsilon_0 x} \\ & =\frac{9 \times 10^9 \times 4 \times 10^{-6}}{2} \\ & =18000 \mathrm{~N} / \mathrm{C} \end{aligned} $$
Electric potential, $V=\frac{2 \times 10^{-6}}{4 \pi \epsilon_0(3-x)}+\frac{(-4) \times 10^{-6}}{4 \pi \epsilon_0 x}$ $$ \begin{aligned} & 0=\frac{2}{3-x}-\frac{4}{x} \\ & \Rightarrow 0=\frac{1}{3-x}-\frac{2}{x} \Rightarrow \frac{1-}{3-x}=\frac{2}{x} \\ & \Rightarrow x=6-2 x \\ & \Rightarrow 3 x=6 ; x=2 m \end{aligned} $$ Electric field at a point $P$ is $$ \begin{aligned} & E=\frac{Q_2}{4 \pi \epsilon_0 x} \\ & =\frac{9 \times 10^9 \times 4 \times 10^{-6}}{2} \\ & =18000 \mathrm{~N} / \mathrm{C} \end{aligned} $$
Hence, the correct answer is option: 'B'.