JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 2)
Two charges $$q$$ and $$3 q$$ are separated by a distance '$$r$$' in air. At a distance $$x$$ from charge $$q$$, the resultant electric field is zero. The value of $$x$$ is :
$$\frac{r}{3(1+\sqrt{3})}$$
$$\frac{(1+\sqrt{3})}{r}$$
$$\frac{r}{(1+\sqrt{3})}$$
$$r(1+\sqrt{3})$$
Explanation
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$$\begin{aligned} & \left(\vec{E}_{\text {net }}\right)_P=0 \\ & \frac{\mathrm{kq}}{\mathrm{x}^2}=\frac{\mathrm{k} \cdot 3 \mathrm{q}}{(\mathrm{r}-\mathrm{x})^2} \\ & (\mathrm{r}-\mathrm{x})^2=3 \mathrm{x}^2 \\ & \mathrm{r}-\mathrm{x}=\sqrt{3} \mathrm{x} \\ & \mathrm{x}=\frac{\mathrm{r}}{\sqrt{3}+1} \end{aligned}$$
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