JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 16)
Force between two point charges $$q_1$$ and $$q_2$$ placed in vacuum at '$$r$$' cm apart is $$F$$. Force between them when placed in a medium having dielectric constant $$K=5$$ at '$$r / 5$$' $$\mathrm{cm}$$ apart will be:
$$5 F$$
$$25 F$$
$$F / 5$$
$$F / 25$$
Explanation
In air $$F=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}_2}$$
In medium $$\mathrm{F}^{\prime}=\frac{1}{4 \pi\left(\mathrm{K} \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{\left(\mathrm{r}^{\prime}\right)^2}=\frac{25}{4 \pi\left(5 \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{(\mathrm{r})^2}=5 \mathrm{~F}$$
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