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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)

  1. A 3 cm
  2. B 2 cm
  3. C 4 cm
  4. D 1 cm
Verified Solution

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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