
Charges of $2 \mu \mathrm{C}$ and $-3 \mu \mathrm{C}$ are placed at two points A and B separated by 1 m .…

- 0.7 m
- 0.5 m
- 0.4 m
- 0.6 m
Solution
Let $r_1$ and $r_2$ be the distances from points $A$ and B respectively where the potential is zero. $\begin{array}{ll} \therefore \quad & \frac{\mathrm{K}\left(2 \times 10^{-6}\right)}{\mathrm{r}_1}=\frac{\mathrm{K}\left(3 \times 10^{-6}\right)}{\mathrm{r}_2} \\ \therefore \quad & \frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{3}{2} \\ \therefore \quad & \frac{\mathrm{r}_2+\mathrm{r}_1}{\mathrm{r}_1}=\frac{2+3}{2} \\ \therefore \quad & \frac{1}{r_1}=\frac{5}{2} \ldots\left(\text { given, } r_1+r_2=1 \mathrm{~m}\right) \\ \therefore \quad & r_1=0.4 \mathrm{~m} \end{array}$ /
Asked in: MHT CET 2024 (15 May Shift 2)