JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 8)
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 $9 \times 10^9$ in SI units)
1 cm
2 cm
4 cm
3 cm
Explanation
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$$\begin{aligned} & \mathrm{F}=\frac{\mathrm{k}\left(\frac{\theta}{2}\right)\left(\frac{\theta}{2}\right)}{\mathrm{r}^2} \\ & 9 \times 10^{-3}=\frac{9 \times 10^9 \times\left(4 \times 10^{-8}\right) \times 4 \times 10^{-8}}{4 \times \mathrm{r}^2} \\ & \mathrm{r}^2=\frac{9 \times 10^9 \times 16 \times 10^{-16}}{4 \times 9 \times 10^{-3}}=4 \times 10^{-4} \\ & \mathrm{r}=2 \times 10^{-2} \mathrm{~m} \Rightarrow 2 \mathrm{~cm} \end{aligned}$$
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