JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2025
- Hard
- MCQ
A small uncharged conducting sphere is placed in contact with an identical sphere but having \(4 \times 10^{-8} \mathrm{C}\) charge and then removed to a distance such that the force of repulsion between them is \(9 \times 10^{-3} \mathrm{~N}\). The distance between them is (Take \(\frac{1}{4 \pi \epsilon_{\mathrm{o}}}\) as \(9 \times 10^9 \mathrm{in} \mathrm{SI}\) units)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 2 cm
Step-by-step Solution
Detailed explanation

\(\begin{aligned} & \mathrm{F}=\frac{\mathrm{k}\left(\frac{\theta}{2}\right)\left(\frac{\theta}{2}\right)}{\mathrm{r}^2} \\ & 9 \times 10^{-3}=\frac{9 \times 10^9 \times\left(4 \times 10^{-8}\right) \times 4 \times 10^{-8}}{4 \times \mathrm{r}^2} \\ & \mathrm{r}^2=\frac{9 \times 10^9 \times 16 \times 10^{-16}}{4 \times 9 \times 10^{-3}}=4 \times 10^{-4} \\ & \mathrm{r}=2 \times 10^{-2} \mathrm{~m} \Rightarrow 2 \mathrm{~cm}\end{aligned}\)
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