JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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)
- A 3 cm
- B 2 cm
- C 4 cm
- D 1 cm
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…
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