MHT CET · Physics · Electrostatics
Charges of \(2 \mu \mathrm{C}\) and \(-3 \mu \mathrm{C}\) are placed at two points A and B separated by 1 m . The distance of the point from A , where net potential is zero, is

- A 0.7 m
- B 0.5 m
- C 0.4 m
- D 0.6 m
Answer & Solution
Correct Answer
(C) 0.4 m
Step-by-step Solution
Detailed explanation
\(V=\frac{K q}{r}\)
Let \(r_1\) and \(r_2\) be the distances from points \(A\) and B respectively where the potential is zero.
\(\begin{array}{ll}
\therefore \quad & \frac{\mathrm{K}\left(2 \times 10^{-6}\right)}{\mathrm{r}_1}=\frac{\mathrm{K}\left(3 \times 10^{-6}\right)}{\mathrm{r}_2} \\
\therefore \quad & \frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{3}{2} \\
\therefore \quad & \frac{\mathrm{r}_2+\mathrm{r}_1}{\mathrm{r}_1}=\frac{2+3}{2} \\
\therefore \quad & \frac{1}{r_1}=\frac{5}{2} \ldots\left(\text { given, } r_1+r_2=1 \mathrm{~m}\right) \\
\therefore \quad & r_1=0.4 \mathrm{~m}
\end{array}\)
Let \(r_1\) and \(r_2\) be the distances from points \(A\) and B respectively where the potential is zero.
\(\begin{array}{ll}
\therefore \quad & \frac{\mathrm{K}\left(2 \times 10^{-6}\right)}{\mathrm{r}_1}=\frac{\mathrm{K}\left(3 \times 10^{-6}\right)}{\mathrm{r}_2} \\
\therefore \quad & \frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{3}{2} \\
\therefore \quad & \frac{\mathrm{r}_2+\mathrm{r}_1}{\mathrm{r}_1}=\frac{2+3}{2} \\
\therefore \quad & \frac{1}{r_1}=\frac{5}{2} \ldots\left(\text { given, } r_1+r_2=1 \mathrm{~m}\right) \\
\therefore \quad & r_1=0.4 \mathrm{~m}
\end{array}\)
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